Let $G$ be a group (finite, say) with center $Z$. The automorphism group $\text{Aut}(G)$ sits in a short exact sequence
$$ 1 \to G/Z \to \text{Aut}(G) \to \text{Out}(G) \to 1. $$
So when $Z\neq 1$, as is often the case, $G$ does not itself naturally embed in its automorphism group.

**Questions:** I would like to know if there is any way to remedy this defect and extend $\text{Aut}(G)$ to contain $G$. [I'm mainly interested in the case where $G$ is an extraspecial $2$-group, but I think the general question is kind of fun.] Specifically:

1. Is there any extension 
$$ 1 \to G \to \widetilde{\text{Aut}}(G) \to \text{Out(G)} \to 1?$$

2. If so, is there a natural construction of such an extension?

I suspect the answer to Question 1 is "no": it feels too good to be true. Even so, it would still be nice to know:

3. What if we replace $\text{Aut}(G)$ by the group $\text{Aut}^\circ(G)$ of automorphisms trivial on $Z$?

4. What is the obstruction to $\widetilde{\text{Aut}}(G)$ existing?

The group $\widetilde{\text{Aut}}(G)$ would be an extension of $\text{Aut}(G)$ by $Z$, so there is some hope of describing it using group cohomology. From this perspective, the question becomes, in a weak form:

5. Must the image of the pullback map $H^2(\text{Aut}(G),Z) \to H^2(G/Z,Z)$ contain a class corresponding to $G$?

**Examples**: These are surely too simple, but give some small positive evidence:

 - If $G=Z$ is abelian then we can take $\widetilde{\text{Aut}}(Z) = Z \rtimes \text{Aut}(Z)$.

- More generally, if the map $\text{Aut}(G) \to \text{Out}(G)$ splits (that is, admits a homomorphic section) then we can take $\widetilde{\text{Aut}}(G) = G \rtimes \text{Out}(G)$. Here $\widetilde{\text{Aut}}(G)$ would seem to depend on the choice of splitting.