I asked this previously in an earlier question, although admittedly my question was hard to find in the slew of info I provided there, so hopefully this will clarify things. Suppose there's a typical $F=ma$ system of the form $$M\ddot{\mathbf{x}}=-\nabla U(\mathbf{x})$$ where $M$ is a mass tensor, $U$ is the potential and $\mathbf{x}$ is a vector describing the state. Suppose that there's a matrix representation of a finite group $G$, $\{\Gamma_g|g\in G\}$, such that $U(\Gamma_g\mathbf{x})=U(\mathbf{x})$ for all $\mathbf{x}\in \mathbb{R}^n$ and $g\in G$. Likewise, let's assume there's a basis $\Lambda=\{\Lambda_1,...,\Lambda_n\}$ for $\mathbb{R}^n$ such that each basis vector is in a definite eigenstate of each $\Gamma_g$.
If we further assume that the system starts out with initial conditions which are in eigenstates of all the $\Gamma_g$, $$\Gamma_g\mathbf{x}(0)=\lambda_g\mathbf{x}(0)\mbox{ and }\Gamma_g\dot{\mathbf{x}}(0)=\lambda_g\dot{\mathbf{x}}(0)$$ then what, if anything, can we say about the expectation values of the symmetry numbers of the state, $$\left<g(t)\right>:=\left<\mathbf{x}(t),\Gamma_g\mathbf{x}(t)\right>$$ as time progresses? When are symmetry numbers conserved over time? If nothing can be said, are there additional constraints which would allow us to say something definite?
Here's what I tried: Without loss of generality we can switch to mass-weighted coordinates to get rid of $M$. I vaguely suspected that if you happened to be able to show that $\ddot{\mathbf{x}}(t)$ had the same symmetry as the position and velocity, then the symmetries would be conserved. We have $$\Gamma_g\ddot{\mathbf{x}}(t)=\nabla U(\Gamma_g \mathbf{x}(t))$$ and since $G$ is finite and $\Gamma_g$ is real, the only possible eigenvalues of $\Gamma_g$ are $1$ or $-1$. In the case where $\mathbf{x}(t)$ is a gerade eigenstate of $\Gamma_g$, we have $$\Gamma_g\ddot{\mathbf{x}}(t)=\nabla U(\mathbf{x}(t))=\ddot{\mathbf{x}}(t)$$ which implies the force is of $g$-gerade symmetry, and I feel like we can conclude in this case that $\left<g(t)\right>=1$ for all $t$. Thus there is a "transition selection rule" which says that the system will evolve in such a way which forces the $g$-ungerade entries of $\Lambda^\mathsf{T}\mathbf{x}(t)$ to always be zero. Is that correct? What about $g$-ungerade states $\mathbf{x}(t)$?
By the way, I never learned Noether's theorem, although I vaguely recall it has something to do with conserved continuous symmetries, so if you feel this can be answered as a special case of Noether's theorem with discrete symmetries, by all means say so :)
