= Solution
Let $U(t,t_0)$ be the unitary Schrödinger propagator. The Heisenberg state is fixed at its initial value, while $A_H=U^\dagger A_SU$. Since $\psi_S(t)=U\psi_H$, matrix elements and probabilities agree exactly in the two pictures. Differentiating, for an operator with no explicit Schrödinger-time dependence, gives
$$
\boxed{\dot A_H=\frac i\hbar[H_H,A_H]}.
$$
For constant <Hamiltonian operator>, $U=e^{-iH(t-t_0)/\hbar}$ and $H_H=H$.
A spatial rotation is represented by $U_R(\theta)=\exp[-i\theta n\cdot J/\hbar]$, where $J$ is the total <angular momentum operator>. For a particle $J=L+S$. The commutators $[J_i,x_j]=i\hbar\epsilon_{ijk}x_k$ and $[J_i,S_j]=i\hbar\epsilon_{ijk}S_k$ give
$$
U_R^\dagger xU_R=x+\theta n\times x+O(\theta^2),\qquad
U_R^\dagger SU_R=S+\theta n\times S+O(\theta^2).
$$
These are the correct infinitesimal vector rotations, verifying the generator and sign.
For the specified <Hamiltonian operator>, $p^2$ and $x^2$ commute with every $J_i$. Moreover
$$
[J_i,x\cdot S]=i\hbar\sum_{j,k}\epsilon_{ijk}(x_kS_j+x_jS_k)=0
$$
by antisymmetry. Thus $[J_i,H]=0$ and \b[every total-angular-momentum component is constant in the <Heisenberg picture>]. The separate torques are
$$
\boxed{\dot L=-U(x^2)x\times S,\qquad \dot S=+U(x^2)x\times S}.
$$
They cancel but need not vanish individually, so orbital and <spin angular momentum> are generally not separately conserved.
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