Subgroup of "equivariant" maps?

Question

Given a $G$-set $X$, I've checked that the set: $$E(G,X,\cdot):=\lbrace f\in\operatorname{Sym}(X)\mid f(g\cdot x)=g\cdot f(x), \text{ for all }g \text{ and all }x\rbrace$$ is a subgroup of $\operatorname{Sym}(X)$. Unless it is always $E(G,X,\cdot)=\operatorname{Sym}(X)$ (possibily for some trivial reason I'm missing), does this subgroup have a name?

Answer 1

It is just the automorphism group of the $G$-set $X$. The concept of automorphism groups makes sense in every category, and here you have the category of $G$-sets. In my opinion, giving these groups special names obscures the fact that they are the same concept (in disguise, when you ignore basic category theory).

The reason that some authors prefer to say "equivariant automorphisms", "linear automorphisms", "bijective isometries" etc. instead of just "automorphisms" is to hide the usual implicit application of forgetful functors (here: from $G$-spaces to spaces, or $G$-sets to sets).