Isomorphisms of infinite graphs that are fixed on a set of vertices

Question

Suppose we have two infinite directed graphs $\langle V,E\rangle$ and $\langle V,E^*\rangle$, and a set $A \subseteq V$ such that, for all finite $X \subseteq A$, there is an isomorphism from $\langle V,E\rangle$ to $\langle V,E^*\rangle$ that fixes $X$ (in the sense of mapping every member of $X$ to itself). Does it follow that there is an isomorphism from $\langle V,E\rangle$ to $\langle V,E^*\rangle$ that fixes $A$?

Answer 1

Counterexample. Let $V=\mathbb Z\cup F$ where $F$ is the set of all functions $f:\mathbb N\to\{1,-1\}$ such that $f(n)=1$ for all sufficiently large $n$. Let $E=\{\langle f,f(n)n\rangle:n\in\mathbb N\}$, $E^*=\{\langle f,-f(n)n\rangle:n\in\mathbb N\}$.

Clearly, there is no isomorphism from $\langle V,E\rangle$ to $\langle V,E^*\rangle$ which fixes $\mathbb Z$. However, for each $m\in\mathbb N$, there is an isomorphism $\varphi:\langle V,E\rangle\to\langle V,E^*\rangle$ which fixes the set $X=\{x\in\mathbb Z:|x|\le m\}$. Namely, for $x\in\mathbb Z$, define $$\varphi(x)=\begin{cases} \ \ \ x\text{ if }|x|\le m,\\ -x\text{ if }|x|\gt m;\\ \end{cases}$$ and for $f\in F$ and $n\in\mathbb N$ define $$\varphi(f)(n)=\begin{cases} -f(n)\text{ if }n\le m,\\ \ \ \ f(n)\text{ if }n\gt m.\\ \end{cases}$$