= Solution
Consider the discrete map
$$
(t,r,\psi,\theta,\phi)\longmapsto
(t,r,\psi,\pi-\theta,\,2\phi_0-\phi).
$$
It is an <isometry>: $\cos\theta$ and $d\phi$ both change sign, leaving $\cos\theta\,d\phi$ unchanged, while $d\theta^2$ and $\sin^2\theta\,d\phi^2$ are also unchanged. A local component of its <fixed-point set> is $\theta=\pi/2$, $\phi=\phi_0$.
If a <geodesic> starts tangent to that component, the <isometry> fixes both its initial position and its initial tangent. Applying the <isometry> therefore gives a <geodesic> with the same initial data. Uniqueness of the <geodesic equation> makes it the same curve, so it remains in the fixed component. This proves that the selected submanifold is a <totally geodesic submanifold>; it has dimension three, not two. The proof continues through a horizon when expressed in regular coordinates, so coordinate singularities of the original chart do not invalidate it.
The <geodesic conserved quantities from Killing vectors> associated with $\partial_\psi$ and $\partial_\phi$ are
$$
L_\psi=r^2h\left(\dot\psi+\frac12\cos\theta\,\dot\phi-\Omega\dot t\right),\qquad
L_\phi=\frac12\cos\theta\,L_\psi+\frac{r^2\sin^2\theta}{4}\dot\phi.
$$
Dots denote <derivatives> with respect to an <affine parameter>. On the chosen submanifold $\dot\phi=0$ and $\cos\theta=0$, so $L_\phi=0$. Requiring also $L_\psi=0$ gives
$$
\boxed{\dot\psi=\Omega(r)\dot t.}
$$
More generally, away from the polar-coordinate axes, both displayed angular charges being zero imply $\dot\phi=0$ and the same relation. In particular, zero conserved <angular momentum> does not imply constant $\psi$: the motion follows the local dragging of the angular coordinate by the rotating geometry.
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