In searching the title of my paper, I accidentally came across this post! In addition to the excellent responses above, you can check this recent paper published in the AIAA Journal of Guidance, Controls and Dynamics (https://arc.aiaa.org/doi/10.2514/1.G007409) in which I have provided answers to your questions and provided analytic bounds on the lower (required) and upper (allowable) number of impulses for three classes of maneuvers:
fixed-terminal-time rendezvous,
free-terminal-time rendezvous,
and phase-free transfer (directly related to Edelbaum's question). I encourage you to first read the "How Many Impulses Redux" paper (https://link.springer.com/article/10.1007/s40295-019-00203-1), since it is a good starting point (in particular for the results regarding the Earth-Dionysus problem), and then continue to read the recent JGCD paper. In case you don't have subscription to JGCD, I have uploaded a version of the JGCD paper before being formatted by JGCD on ResearchGate (https://www.researchgate.net/publication/372525067_Existence_of_Infinitely_Many_Optimal_Iso-Impulse_Trajectories_in_Two-Body_Dynamics)
In summary and considering the simplest example possible, the minimum number of impulses is one. Consider GTO and GEO, where the apogee of the GTO is tangent to the GEO. You can also assume that GTO has a non-zero inclination value. If the starting point of the maneuver is perigee of the GTO, at precisely half of the orbital period of the GTO, the spacecraft reaches the apogee point and one single maneuver is needed to optimally inject the spacecraft onto the GEO. This is the case in which there is one intersecting point between the two orbits. I coined the term "impulse anchor position" since there are cases that there are two impulse anchor positions (check Earth-Dionysus problem in the JGCD paper).
How about the maximum number of impulses, which appears to be a far more complex problem? However, it is possible to divide the single delta-v vector at the intersection point (impulse anchor position) into many small-magnitude, same-direction impulses such that the total sum of the smaller impulses is still equivalent to the magnitude of the original impulse. The only inevitable change to the problem is that we need extra time. In principle, any minimum-deltav (respecting Lawden's primer vector theory) maneuver can be broken down into 4 main phases: 1) motion on the original orbit, 2) motion on an unknown number of phasing orbits, 3) motion on a connecting arc (associated with a phase-free transfer between impulse anchor positions) and 4) motion on the target orbit. Depending on the available time, the spacecraft can make multiple revolutions (around the central body) on these 4 main phases. Regarding the GTO-GEO problem, each small impulse, places the spacecraft on an intermediate elliptical phasing orbit. Thus, if we are interested in a transfer maneuver (and if the time has no bound), you can have infinitely many small-magnitude impulses that achieve the transfer (i.e., move the spacecraft from one orbit into another). So, the answer to your question is infinity! Note that if the time of flight is finite/bounded, then, the maximum number of impulses is finite.
For all infinity of solutions, the total delta-v is equal to the single-impulse optimal one, and we have essentially stretched the maneuver time without sacrificing optimality with respect to the delta-v. Why do we want to stretch the time? Well, if the propulsion system of the spacecraft can only produce a maximum value of impulse, with this flexibility, we can throw as many phasing orbits as needed to ensure that the largest impulse is less than the maximum value (threshold). So, this flexibility is significant from an operational point of view at the cost of increasing the mission time, but without incurring any additional cost on delta-v, namely, delta-v optimality is retained.
On the other hand, if the problem is a rendezvous-type maneuver and the time of flight has an upper bound, the highest number of impulses is always finite. How can we determine the maximum number of impulses? We can use the available time of flight and use the lowest orbital period (among the 4 phases of motion) and analytically obtain the maximum number of phasing orbits that can be added to the motion and also determine the maximum number of impulses (again, check the Earth-Dionysus problem in the link to the JGCD paper on (ResearhGate)). More examples are given in the paper (with entire solution envelops for possible ranges of time spent on each phase of motion), but the entire idea is based on the key observations that I outlined for the GTO-GEO problem.
Updates to the response regarding the "required" versus "could" number of impulses. My response is quite lengthy to be included as a comment and I had to update my response.
In orbital mechanics/astrodynamics, the class of transfer maneuvers is characterized in a specific manner, i.e., it corresponds to time-free, phase-free maneuvers that make it possible for a spacecraft to move from one orbit and to be inserted into another orbit. Since according to the definition both time and phase are free, orbit transfer maneuvers are the least constrained type of trajectories that are typically used for gaining additional insights into the theoretically optimal (with respect to delta-v) solutions. With this definition in mind, and as I explained earlier, infinitely many impulses are the answer to the highest number of impulses.
Allow me to provide more explanations to clarify my point above. The original question posed by Edelbaum in his 1966 paper is raised within the context of inverse-square restricted two-body dynamics. In principle, we are seeking solutions to a nonlinear dynamical system. Potential solutions to nonlinear dynamical systems may (and frequently do) exhibit bifurcations with respect to both temporal component (i.e., time) and other parameters of the problem. It is important and crucial to take both of these components into consideration when we are interested in analyzing complex nonlinear dynamical systems.
For example, consider the problem of maneuvering a spacecraft from a planer circular LEO to another large-raidus circular orbit. It is a classic result that if the ratio of radii of the larger orbit and LEO is less than 11.94, then, the Hohmann maneuver is the most efficient transfer maneuver. But, in this result, it is implied that time is unbounded since by definition we are dealing with transfer maneuvers. Question: If the ratio of radii is less than 11.94, how many impulses are required to perform an optimal (in the sense of delta-v) orbital transfer? The answer depends on the value that we choose/allow for the time of flight. If time of flight is constrained to be less than half of the orbital period of the Hohmann ellipse (T_Hoh), then, we have no option, but to solve a time-constrained Lambert problem, which results in a bi-impulsive maneuver. We can find the best (least delta-v) bi-impulsive maneuver, but the solution will never be optimal from a delta-v perspective. However, if the time of flight is exactly equal to T_Hoh, then, the optimal number of impulses is two and we can guarantee optimality of the solution (w.r.t. delta-v). If the time of flight is constrained to be greater than T_Hoh, then, we have to split the impulses at the anchor positions to match the time of flight without sacrificing optimality in delta-v. So, we may need more than two impulses to retain delta-v optimality. From these three cases, it is clear that we have to relax any constraint on the time of flight to attain the optimal solution. Thus, for the range of the ratio parameter of the radii, the minimum number of impulses is two (corresponding to the classic Hohmann transfer), whereas the maximum number of impulses is infinity and all of these are optimal with respect to delta-v. But there is more to these solutions than meets the eye!
Another example will shed more light on what I explained above. What if in the above example the ratio of the radii is between 11.94 and 15.58? Again, it is theoretically proven that we can have Hohman and Bielliptic maneuvers that have the same total delta-v value, i.e., delta-v_Hoh = delta-v_BE. In other words, there is a separatrix between Hohmann and Bielliptic maneuvers when the ratio is between 11.94 and 15.58. The key point is that time plays a significant role in the (required) number of optimal impulses. If we relax/ignore time, again, there is an infinity of two- and three-impulse solutions (corresponding to the separatrix curve) that are optimal with respect to delta-v. If we fix the ratio of the radii (between 11.94 and 15.58), as I explained, we can have unique two-impulse (Hohmann) and three-impulse (Bielliptic) maneuvers that have exactly the same delta-v. So, the number of impulses is not unique. The difference between the two solutions is the time of flight. Check chapter six of either third or fourth editions of the Book by Howard D. Curtis (Fig 6.8 in the Third/Fourth editions) for a plot that summarizes the results and also depicts the separatrix (corresponding to the equal-delta-v curve).
As is evident, time and other components (e.g., ratio of the radii) can result in different classes of solutions. Characteristics of the solutions differ according to these values. Note also that in dynamical systems (while often ignored), feasibility is a more fundamental criterion than pure optimality. In the circular-to-circular example with a ratio of radii less than 11.94, if the imposed time of flight is less than T_Hoh, we cannot expect any optimality since time being less than T_Hoh restricts the set of feasible solutions to the class of two-impulse Lambert solutions. We can find the best (with respect to delta-v) two-impulse solution, but that solution is sub-optimal compared to the delta-v of the Hohmann maneuver. That is why we have to relax the time of flight to enlarge the class of feasible solutions, which is the idea in studying transfer maneuvers.
Without going into the details, the same analysis is true for other problems. In the JGCD paper, I have shown that for the Earth-Dionysus problem, if we limit the time of flight to be fixed at 9.7 years, 139 different classes of equal-delta-v solutions exist. The lowest number of required impulses is three and the largest number of required impulses is 8. All of these solutions require the same amount of delta-v. In addition, I have shown that if the time of flight is constrained to be less than 9.7 years, a natural consequence is reduction in the set of feasible equal-delta-v solutions. However, there is a specific time of flight below which we have to abandon delta-v optimality. To address the time-feasibility problem of the Earth-Dionysus problem, I have mentioned that if the time of flight is less than 1840.31 days, it is impossible to obtain any optimal solution (with respect to delta-v), and as such, a two-impulse solution is the only option; however, the delta-v corresponding to the bi-impulsive maneuver is never going to be less than its value for the three-impulse maneuver. In other words, we have proposed an algorithm that allows us to determine time-feasibility of a minimum-delta-v maneuver.
In summary, when we are interested in determining the maximum number of required impulses, time-feasibility has to be considered and a constraint on the time of flight has to be imposed. Below a certan time, it is impossible/infeasible to obtain any theoretically optimal solution. There are ranges of the time parameter for which the number of impulses can change to retain optimality. For transfer maneuvers, the maximum number of impulses is infinity since time has no bound by definition. You can check the details of the Earth-Dionysus paper in the JGCD paper.
Best, Ehsan Taheri