The present question follows up this one, in which I accidentally asked for less than I actually wanted. Given a group $G$, I would like to find an extension $\tilde G$ of its automorphism group $\operatorname{Aut}G$ by its center $Z(G)$, into which $G$ embeds in such a way that the restriction of the surjection $\tilde G\rightarrow\operatorname{Aut}G$ to $G$ maps $g\mapsto \rho_g$, where $\rho_g$ is conjugation by $g$.
With a little more precision: starting from a group $G$, I want an exact sequence
$$1\rightarrow Z(G) \rightarrow\tilde G\xrightarrow{\varphi} \operatorname{Aut}G\rightarrow 1$$
and an embedding $i:G\hookrightarrow \tilde G$, such that $\varphi|_{i(G)}$ is the map $i(g)\mapsto \rho_g$.
My question is this:
Does the desired $\tilde G$ always exist, and for what $G$'s is it unique? Is it always unique?
I am happy to assume that $G$ is finite if that is helpful.
Thanks in advance.
Asides:
(1) I think that finding my desired $\tilde G$ is equivalent to finding an element of $H^2(\operatorname{Aut}G,Z(G))$ whose restriction to $H^2(\operatorname{Inn}G,Z(G))$ is the class of $G$.
(2) the difference between this question and the previous one I asked (which was correctly answered by the holomorph) is that here I am requiring that the kernel of the map $\tilde G\rightarrow \operatorname{Aut}G$ be exactly $Z(G)$.
No, such an extension does not exist for $G = {\rm SL}(2,9)$, for example (but I don't know whether that is the smallest such example). In the notation of the ATLAS of Finite Groups, $G = 2.A_6$ is a perfect central extension of $A_6 \cong {\rm PSL}(2,9)$, and ${\rm Aut}(G) = {\rm P \Gamma L}(2,9) = A_6.2^2$.
There are three extensions of $A_6$ by automorphisms of order $2$, namely $S_6$, ${\rm PGL}(2,9)$ and $M_{10}$. The combined extensions $2.S_6$ and $2.{\rm PGL}(2,9)$ both exist, but $2.M_{10}$ does not and hence neither does $2.A_6.2^2$. There is a lot of information about this question for simple groups in the ATLAS.
More generally, for any odd prime $p$, the extension $2.{\rm PSL}(2,p^2).2^2$ does not exist.