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.

Trivially, if $F\subseteq V^V$, the empty set $E = \emptyset$ is the smallest $F$-compatible set. Is there always a largest $F$-compatible set (containing all $F$-compatible sets)?