 Solution

The orbiting planet produces a potential $V(r,\vartheta ,\phi -\Omega t)$. It has explicit time dependence and exerts an azimuthal torque, so generally $\dot{E}=V_t$ and ${\dot{L}}_z=-V_{\phi }$ are nonzero. Nevertheless a simultaneous time-translation symmetry $\delta t=\lambda$ and rotation $\delta \phi =\Omega \lambda$ leaves the Lagrangian and $dt$ unchanged. Noether's theorem gives $Q=\Omega p_{\phi }-E$ and henceEquivalently $V_t=-\Omega V_{\phi }$ implies $d(E-\Omega L_z)/dt=0$. This Jacobi integral assumes a rigidly rotating potential with constant $\Omega$ and negligible back-reaction from the test particle.