 Solution

Stokes flow obeys $-\nabla p+\mu {\nabla }^{2}u=0$, $\nabla \cdot u=0$. Reversing all applied forces reverses $u$ while leaving streamlines unchanged. Reflection in the vertical plane containing the cylinder axis rules out rotation normal to that plane; reflection combined with reversibility rules out rotation within it and spin about the axis. Thus its orientation remains fixed.