 Solution

Reflect the configuration in the horizontal plane through the cylinder axis. The wall and cylinder geometry are unchanged. Downward translation becomes upward translation, a possible wall-normal migration velocity is unchanged, and the axial angular velocity changes sign because it is an axial vector.

On the other hand, reversing the original Stokes flow makes all three velocities change sign: translation becomes upward, wall-normal migration reverses, and rotation reverses. The reflected and reversed problems therefore have the same forcing and geometry except that their wall-normal velocities have opposite signs. Uniqueness of Stokes flow forces that velocity to equal its own negative, so it is zero. The rotational velocities already agree under the two operations, so this argument places no restriction on rotation.