This is an exercise from "Cellular Automata and Groups" by T. Ceccherini-Silberstein and M. Coornaert.
Let $G$ be a locally finite group and let $A$ be an arbitrary set. Let $\tau: A^G \rightarrow A^G$ be a bijective cellular automata ie
(1) that $\tau$ is $G$-equivariant under the action of $G$ on $A^G$ given by $g(x(h)) = x(hg^{-1})$, so, $g\tau(x) = \tau(gx)$ and
(2) that $\tau$ is uniformly continuous with respect to the prodiscrete uniform space equipped to $A^G$.
(or equivalently $\tau$ is $G$-equivariant and there exists finite "memory set" $S \subset G$ and a map $\mu :A^{s} \rightarrow A$ such that $\tau(x)(g) = \mu(g^{-1}x|_{S})$ for the group action above.
I am trying to show that it must be that $\tau^{-1}$ is also a cellular automata. But I am not sure how $G$ being locally finite gets us there. I do have a couple of observations: first, we can assume $A$ is infinite, since the finite case is trivial. Second is that if we let $S$ be the memory set of $\tau$, then $H = \langle S \rangle$ is finite, and I know that $\tau$ is invertible iff the restriction of $\tau$ to $H$ is invertible, and it might be that it is easier to work with $A^H$ with $H$ finite (showing that $\tau^{-1}$ is uniformly continuous) or maybe we get nice some property forced on $A^H$ as a prodiscrete topological or prodiscrete uniform space.
What am I missing?