Let $V$ be a set and let $V^V$ denote the set of all functions $f:V\to V$. Suppose that $F\subseteq V^V$. Let $[V]^2 = \big\{\{x,y\}: x, y\in V \land x\neq y\big\}$. We say $E\subseteq [V]^2$ is *$F$-compatible* if all members of $F$ are graph homomorphisms from $(V,E)$ to itself.

Suppose $G=(V,E)$ is a simple, undirected graph. With $\text{Hom}(G,G)$ we denote the set of all graph homomorphisms $f:V\to V$.

This post says that there is a largest $\text{Hom}(G,G)$-compatible set $E'\subseteq [V]^2$. Clearly $E'\supseteq E$.

**Question.** Do we have $E'=E$?