What symmetry does conserved $L^2$ imply? According to Noether's theorem, if $[L,H]=0$ the system has rotational invariance. Does $[L^2,H]=0$ also imply some symmetry for the system?
 A: By the derivative rule
$$
[H,L^2]=[H,L]L+L[H,L]
$$
so there is noting new in $[H,L^2]$ that is not in $[H,L]$.

Edit: in fact my answer is not correct.  Take $H=L_x$.  Then $[L^2,L_x]=0$ but $[L,L_x]\ne 0$. 
So: if $[L,H]=0$ then $[L^2,H]=0$ implies nothing more, but it could be that $[L^2,H]=0$ without $[L,H]=0$.
A: The symmetry, for a particle described by position $x_k$ and momentum $p_j$ is the one defined by, for $k=1,2,3$
 $$x_k \to x_k(a) := U_a x_k U_a^{-1}$$ 
$$p_k \to p_k(a) := U_a p_k U_a^{-1}$$
where $U_a := e^{iaL^2}$. The formal solution is
$$\vec{x}(a) = \sum_{n=0}^{+\infty} \frac{i^na^n}{n!}[L^2[L^2\cdots (n \:times)\cdots  [L^2,\vec{x}]\cdots]]$$
and
$$\vec{p}(a) = \sum_{n=0}^{+\infty} \frac{i^na^n}{n!}[L^2[L^2\cdots (n \:times)\cdots  [L^2,\vec{p}]\cdots]]$$
where we know that (if my computations are correct)
 $$[L^2, \vec{x}] = 2i \vec{x}\wedge \vec{L} +2 \vec{x}$$
and $$[L^2, \vec{p}] = 2i \vec{p}\wedge \vec{L} +2 \vec{p}$$
I am not sure it is possible to sum these series into closed formulas.
