It is a very known fact that for every group $G$ and an automorphism $\sigma$, the set of fixed points of $\sigma$ is a subgroup of $G$. My question is about the converse.
For a finite group $G$ and a subgroup $H$ of $G$, under which conditions, there exists an automorphism $\sigma$ such that the set of fixed points of $\sigma$ is equal to $H$?
A necessary condition is that $H $ be closed under any function $f $ that is equivariant under automorphisms $\sigma $ of $G $: $f(\sigma (x))=\sigma (f (x)) $ (and similarly for functions of several variables).
For instance, taking inverses, products, and $n $th roots. For the latter see also http://groupprops.subwiki.org/w/index.php?title=Fixed-point_subgroup_of_a_subgroup_of_the_automorphism_group#Weaker_properties
