Group action of $G<\mathbb Z^\infty_2$ over the Golden mean shif I'm am looking for an action of an infinite subgroup of $\mathbb Z^\infty_2$ over the golden mean shift space $$X=\{x\in \{0,1\}^\mathbb N : x_i=1\Rightarrow x_{i+1}=0\}$$ such that any element of $G$ changes only a finite number of positions of elements of $X$. 
 A: Let $F:X\to X$ be the map that replaces an occurrence of $010$ around the origin with $000$ and vice versa. That is,
\[
   F x \triangleq
      \begin{cases}
         \cdots x_{-3}x_{-2}000x_1x_2\cdots\quad &\text{if $x_{-1}x_0x_1=010$,}\\
         \cdots x_{-3}x_{-2}010x_1x_2\cdots &\text{if $x_{-1}x_0x_1=000$,}\\
         x &\text{otherwise.}
      \end{cases}
\]
Note that $F$ is an involution, hence acting like $\mathbb{Z}_2$.  So are its compositions with shifts $F\sigma^i$.  Moreover, $F\sigma^i$ and $F\sigma^j$ commute as long as $|i-j|\geq 3$.  Let us use the shorthand $F_k=F\sigma^{3k}$.
For $k_1,k_2,...,k_n\in\mathbb{Z}$, we get a faithful action of $\mathbb{Z}_2^n$ on $X$ by composing $F_{k_1},F_{k_2},\ldots,F_{k_n}$:
\[
   \Phi^{(b_1b_2\cdots b_n)}x \triangleq
      F_{k_1}^{b_1} F_{k_2}^{b_2}\cdots F_{k_n}^{b_n} x \;.
\]
Similarly, if $\mathbb{G}$ denotes the subgroup of $\mathbb{Z}_2^{\mathbb{Z}}$ consisting of all the sequences with finitely many non-zero elements, we get a faithful action of $\mathbb{G}$ with
\[
   \Phi^b x \triangleq (\prod_{k\in\mathbb{Z}} F_k^{b_k})x \;,
\]
where $\prod_{k\in\mathbb{Z}}$ denotes composition.
