Mathematically, it is obvious that the total orbital angular momentum $L^2$ commutes with the spin-orbit Hamiltonian $\propto\boldsymbol{L}\cdot\boldsymbol{S}$. However, is there an intuitive physical reason for this?
For example, the total angular momentum $J^2$ must commute because there is no external torque, and the total spin $S^2$ must commute because the spin of the electron is constant, but I can't think of any similar argument for $L^2$.
If you are considering energy scales where the $L$ and $S$ quantum numbers still make sense and can be used as the unperturbed basis, then there is not enough energy for the electron to change sub-orbital and hence change its angular momentum $\ell$ (in this case equal to the total angular momentum $L$ as you are considering a single-particle system).
Yes. $\vec L\cdot \vec S$ expands as a sum containing $L_x$, $L_y$ and $L_z$, and $L^2$ commutes with all of these individually so commutes with a sum of these, as it would commute with $\hat n\cdot \vec L$.
One can (in a rather hand-waving way) think of the linear combination of $L_k$ in $\vec S\cdot\vec L$ as defining a “direction” along $\vec S$ (just as $\hat n\cdot \vec L$ is in the direction of $\hat n$) and thus just reorienting the quantization axis. Thus if $\vec S$ is “along” $\hat y$ then so is $\vec L$ and one would the redo angular momentum theory using $\hat y$ as the quantization axis.
