Suppose $G$ acts on a set $X$ and $Y$ and let $f:X \to Y$ be a equivariant map.
We immediately have that $$\text{stab}_{G}(x) \subseteq \text{stab}_{G}(f(x)) $$ as if $g \in \text{stab}_{G}(x)$ then $f(x) = f(gx) = g f(x)$ so $g \in \text{stab}_{G}(f(x))$.
If we restrict the action to a group $H \subseteq G$ we have that $$\text{stab}_{H}(x) = H \cap \text{stab}_{G}(x).$$
My question is: How would one find the largest subgroup $H$ such that $$\text{stab}_{H}(x) \trianglelefteq \text{stab}_{H}(f(x))$$ for all $x \in X$?
Clearly if $H=Z(G)$ the condition holds as then the stabilizers become commutative and the subgroup condition proven earlier implies normality.