Here $L^2$ is defined as $$ L^2=L_x^2+L_y^2+L_z^2 $$ representing the observable of the magnitude of the angular momentum.
There are a lot of proofs showing the $z$-projection of the angular momentum, $L_z$, is conserved upon rotating the $xy$-plane. The proof typically uses the technic to rotate the coordinate by a tiny angle $\epsilon$, and Taylor expand the wave function to the first order, and then require the time independent Schrodinger equation to remain the same, e.g. this proof.
Intuitively, $L_z$ is not conserved under arbitrary rotation, but $L^2$ should.
Although rare, I can find some proof on the Internet, using spherical coordinate and concrete Hamiltonian for hydrogen atom, e.g. proof on page 2 of this node. However, this proof depends on the specific form of the Hamiltonian under certain choice of coordinate.
My question is, assuming rotational symmetry only, without assuming any particular form of the Hamiltonian, can we prove that $L^2$ is conserved?
I tried to mimic the proof of the conservation of $L_z$ to Taylor expand the 3D tiny rotated wave function to second order, since $L^2$ involves terms like $\partial_x^2$, but the computation is too complicated and looks not correct.