Let $A$ be a finite alphabet (just some finite set), $G$ a group and $X\subseteq A^G$ some closed subshift, i.e. a closed subset of $A^G$ invariant with respect to the shift action of $G$. For $x,y\in X$ denote by $\Delta(x,y)$ the set where their supports differ, i.e. $\Delta(x,y):=\{g\in G\colon x(g)\neq y(g)\}$. Finally, let $\sim$ denote the homoclinic equivalence relation on $X$. That is, $x\sim y$ if $\Delta(x,y)$ is finite.
When $X$ is the full shift, $x\sim y\in X$ and $|\Delta(x,y)|=n$, then we can find a sequence $x_0=x,x_1,\ldots,x_n=y\in X$ such that $|\Delta(x_i,x_{i+1})|=1$, for every $i<n$. That is, we can find a path from $x$ to $y$ where on each step we change the configuration just on one element.
This is unlikely to work on more general subshifts. I wonder if there is some (reasonable) class of subshifts (probably necessarily of finite type) such that there is a constant $K$ so that for every $x\sim y\in X$ we can find a sequence $x_0=x,x_1,\ldots,x_n=y\in X$ such that $|\Delta(x_i,x_{i+1})|\leq K$, for every $i<n$. I was thinking about irreducible subshifts of finite type, without success though.
If $X$ is a group shift (meaning that $A$ is a finite group and $X$ happens to be a subgroup of $A^G$) and the subgroup $\Delta_X \subset X$ of homoclinic points is generated by translates of a finite set $S \subset X$ of configurations in the sense that $$ \Delta_X = \langle \{ g \cdot x : g \in G, x \in S \} \rangle, \tag{1}$$ then $X$ has the desired property. This is because $x \sim y$ is equivalent to $x^{-1} y \in \Delta_X$, which is a product of translates of elements of $S$. If $G$ is polycyclic, then every group shift $X \subset A^G$ is necessarily of finite type and has a finite set $S$ with (1).