This question follows up a question I asked on math.SE. This is a refinement and a reference request.

For what groups $G$ does there exist a $Z(G)$-extension of $\operatorname{Aut}G$ (call it $\tilde G$) that contains $G$ normally, in such a way that $\tilde G\rightarrow \operatorname{Aut}G$ describes $\tilde G$'s conjugation action on $G$?

Do you know if this question has been studied? If so, can you point me to references?

**Comments**

It is trivial that the class $\mathscr{C}$ of groups admitting such an embedding includes centerless groups (take $\tilde G = \operatorname{Aut} G$).

Also trivially, $G\in\mathscr{C}$ if $\operatorname{Out}G = 1$. Take $\tilde G = G$.

Also easily, $G\in\mathscr{C}$ if $G$ is abelian. Take $\tilde G$ to be $G$'s holomorph.

It seems to me that also $G\in\mathscr{C}$ if $G$ has any faithful irreducible representation of a unique dimension and $\operatorname{Aut}G$ has trivial Schur multiplier, for the following reason. Let $\zeta:G\rightarrow GL(V)$ be the representation in question. $\operatorname{Aut}G$ acts on $G$'s representations but must fix $\zeta$ since it is the unique irreducible representation of its dimension; so any automorphism of $G$ is induced by conjugation by an element of $GL(V)$, which is determined up to a scalar factor; this gives us a faithful projective representation of $\operatorname{Aut}G$ on $V$, which lifts to an ordinary representation $\xi$ because the Schur multiplier of $\operatorname{Aut}G$ is trivial. Then it seems to me that the subgroup of $GL(V)$ generated by the images of $\zeta$ and $\xi$ can be taken to be $\tilde G$.

Actually in this construction (when $G$ has a faithful irreducible representation $\zeta$ of a unique dimension), the assumption of trivial Schur multiplier for $\operatorname{Aut} G$ is overly restrictive. All we need is that the particular projective representation of $\operatorname{Aut} G$ arising from considering its action on the image of $\zeta$ (as above) represents the trivial class in the Schur multiplier. I am not sure how to tell when this happens.

Even this latter condition seems too much to ask for this construction to work, because it is actually asking for $\tilde G \rightarrow\operatorname{Aut}G$ to split. It seems to me that the construction will work as long as the class of $H^2(\operatorname{Aut}G,\mathbb{C}^\times)$ corresponding to the projective representation of $\operatorname{Aut}G$ lies in the subgroup $H^2(\operatorname{Aut}G, \zeta(Z(G)))$. Again, I am not sure how to tell when this happens.