That's a very expensive scenario in general, and especially with your constraint of 'a few days'. And in your case it's best performed using three burns - the maneuver of orbital plane change performed near distant apoapsis.
First, let's adjust the parameters - since the perigee is at 1000km, it means you used altitudes, not radii (or your satellite would fly underground), so let's adjust that by average Earth radius (6371km).
Pe = 7371km
Ap = 16371km
a = (Ap+Pe)/2 = 11871km (let's mark this semi-major axis as a0)
Geometric proof that 2a = Ap+Pe (accounting for planet radii):
Let's have a look at what our orbits should look like:
The direct orbit change there would be a 90 degrees turn, that's original speed times $\sqrt{2}$
geometric proof: diagonal of a square:
Using Vis Viva equation
$$ v = \sqrt{GM ( {2 \over a} - {1 \over r}) } $$
we can find speed at the apogee: sqrt(G (Mass of Earth) ( 2 / 11871km - 1 / 16371km)) = 3888m/s. Multiplied by the aforementioned square root of 2, our delta-V is 5498m/s. That's way too expensive to do in a single launch.
The moon flyby scenario is right out due to timing constraints and inclinations that aim nowhere towards the moon.
1000x10000 is not a great eccentricity but let's see what we can work out with degrading the (actually optimal) scenario of 'plane change at infinity' to fit our timing constraints. I'll assume the 'few' days to be 10.
For the orbital period to be 10 days, semi-major axis (a1) must be 196060km. Pe+Ap = 2a, so Ap = 2*196060-7371 = 384749km.
- So, first, a burn prograde at intersection of the orbital planes near periapsis to raise apoapsis. The initial speed won't be much different from the periapsis speed, so I'll approximate it with it.
Initial speed, Vis Viva with r=Pe, a=a0: 8636m/s
Final speed, r=Pe, a=a1: 10301m/s
$\Delta v_1$ = 10301 - 8636 = 1665m/s.
- At the intersection of the target orbital plane (which should happen very near the new apoapsis) perform a normal burn to adjust your orbital plane. Our apoapsis speed will be 197m/s Times $\sqrt{2}$ for the 90 degrees turn, $\Delta v_2$ = 279m/s.
The actual orbit may be a little higher or lower to make the arrival provide the correct true anomaly - make the period be the nearest (n+0.5) times the period of the satellite, instead of 10 days sharp. The change in delta-V will be quite small.
- Another 1.664km/s burn ($\Delta v_3$), retrograde, at periapsis to bring apoapsis back.
A total of 2*1664+279 = 3607m/s and that's about the least delta-V you can get it done with.
In reality, this will be done through two separate launches, because it will be cheaper than using a launcher that can bring up a transfer stage of over 3.5km/s and three reignitions besides the satellites.
The whole procedure would look kinda like this:
