Jupiter's orbital velocity ranges from 12.4 km/s to 13.7 km/s. To get heliocentric retrograde from a single Jupiter flyby you would need to steer your flyby departing asymptote opposite of Jupiter's motion w.r.t. the Sun (w/ $v_{\infty}$ greater than Jupiter's orbital velocity).
In 2023's Jupiter-direct launch window, trajectories exist with conceivably feasible launch C3 requirements:

(Personal work, filtered for $C3 < 200$ $km^2/s^2$ & $v_{\infty,J} > 12.4$ $km/s$)

(Personal work, individual trajectories from above)
The remaining question is if it's possible to deflect the trajectory that much at Jupiter. At an unrealistic flyby radius of 1 $R_J$ deflection angles are over 100°, and at a more reasonable 10 $R_J$ deflection angles are a considerable ~60°:

(Personal work, flyby deflection angle in degrees (y-axis), flyby $v_{\infty}$ (x-axis), Desmos graph link)
Using the required deflection angle (from the incoming $\vec{v}_{\infty}$ determined by the Earth-Jupiter transfer orbit) and the $v_{\infty}$ magnitude to calculate the flyby close approach distance we get this:
$$r_p = \frac{\mu}{{v_{\infty}}^2} \left( \frac{1}{sin(\frac{\delta}{2})} -1 \right)$$

(Personal work, individual trajectories)
This shows that a single, very close, flyby of Jupiter can put a spacecraft into a retrograde orbit of the Sun.
Here is an animation of what an example trajectory from this batch looks like:
(Personal work)
Further Details of this specific trajectory:
- Launch Date: 23-Jul-2023
- Flyby Date: 09-Jul-2024 (352 day transfer time-of-flight)
- Launch C3: 188 $km^2/s^2$, Flyby $v_{\infty}$: 19.5 km/s
- Flyby Deflection: 77.8°, Flyby Close Approach: 196,500 km (2.75 $R_J$)
- Post Flyby Heliocentric Orbit:
- SMA: 2.8 AU
- Eccentricity: 0.79
- Inclination: 178.7° (retrograde)