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 $X$. 



Group action of $G<\mathbb Z^\infty_2$ over the Golden mean shif
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.


