My original answer misread the question and so the answer is orthogonal to what you are actually asking. As it had some up-votes, I will keep it deleted.
Your proposed statement is false. Here's a counterexample:
Let $G=\mathbb{Z}$. Then there is no group $A$ such that $\mathrm{Inn}(A)\cong\mathrm{Aut}(G)\cong C_2$. In particular, there can be no group $A'$ satisfying the conditions you give.
That no such $A$ exists follows from the following well-known proposition:
Proposition. Let $G$ a group, and let $N\leq Z(G)$. If $G/N$ is cyclic, then $G$ is abelian.
This has been proven many times
in
this
site.
In particular, $\mathrm{Inn}(A)$ cannot be cyclic and nontrivial, because $\mathrm{Inn}(A)\cong G/Z(G)$; this is the case in the example above, since $\mathrm{Aut}(\mathbb{Z})$ is cyclic of order $2$. Thus, there is no group $A$ such that $\mathrm{Inn}(A)\cong\mathrm{Aut}(\mathbb{Z})$. In particular, there can be no group $A$ that contains a normal copy of $\mathbb{Z}$ and satisfies $\mathrm{Inn}(A)\cong\mathrm{Aut}(\mathbb{Z})$.
Likewise, using primitive roots, we have that $\mathrm{Aut}(\mathbb{Z}_n)$ is cyclic for any $n$ that is an odd prime power, twice an odd prime power, $n=2$, and $n=4$; for all those values of $n$ except for $n=2$, there can be no $A$ with $\mathrm{Inn}(A)\cong \mathrm{Aut}(\mathbb{Z}_n)$. In particular, there is no group $A$ that contains a copy of $\mathbb{Z}_n$ as a normal subgroup and has inner automorphism group equal to the automorphism group of $\mathbb{Z}_n$. So the proposition cannot be rescued even if we restrict to finite groups.
On the other hand, Takao Matsumoto proved that every group is isomorphic to the outer automorphism group of some group (Any group is represented by an outerautomorphism group.
Hiroshima Math. J. 19 (1989), no. 1, 209–219, MR 1009671 (90g:20051)), and more recently it was proven that
every group is the outer automorphism group of a simple group.
As to your initial paragraph: if $G$ is centerless, then you can certainly take $A=\mathrm{Aut}(G)$; then $G\cong \mathrm{Inn}(G)\triangleleft \mathrm{Aut}(G)$, and the elements not in $G$ act by conjugation on $\mathrm{Inn}(G)$ inducing an automorphism of $G$ that is not inner.
Let's work with the example of the simple group $ A_6 $ of order $ 360 $.
The theory of quasisimple groups is nice in the sense that "everything you can think to ask for exists." The Schur multiplier of $ A_6 $ is cyclic $ 6 $. So you can ask for the quasisimple groups $ 2.A_6,3.A_6,6.A_6 $ and they all must exist. Fun fact: $ 2.A_6 $ is better known as $ SL(2,9) $.
The theory of almost simple groups is another situation which is nice again in the sense that "everything you can think to ask for exists." $ Out(A_6) $ is $ 2 \times 2 $. So there are four nontrivial subgroups of $ Out(A_6) $: the whole group $ 2 \times 2 $, and three different groups of order $ 2 $ generated by the three different outer automorphisms of order $ 2 $. Again everything we can think to ask for exists: the whole group $ 2 \times 2 = Out(A_6) $ corresponds to $ Aut(A_6) $, while the three outer automorphisms of order $ 2 $ correspond to $ S_6, PGL(2,9) $ and $ M_{10} $ respectively. Fun fact: $ Aut(A_6) $, and thus also the subgroups $ A_6, S_6, PGL(2,9), M_{10} $, all have nice permutation representations of degree $ 12 $.
The theory of almost quasisimple groups, however, is very subtle and not so nice in the sense that we are not guaranteed that everything we can think to ask for exists. Indeed many things we can think to ask for do not exist. Take the case of $ A_6 $. Of course $ 2.A_6,3.A_6,6.A_6 $ all exist. We can also think to ask for almost quasisimple groups of the form $ 2.S_6, 3.S_6, 6.S_6 $, luckily these all exist. We can also think to ask for almost quasisimple groups of the form $ 2.PGL(2,9), 3.PGL(2,9), 6.PGL(2,9) $ again these all exist. However, we can also think to ask for almost quasisimple groups of the form $ 2.M_{10}, 3.M_{10}, 6.M_{10} $. Of these, $ 3.M_{10} $ exists but the other two do not exist. Finally, we can think to ask for almost quasisimple groups of the form $ 2.Aut(A_6), 3.Aut(A_6), 6.Aut(A_6) $. Again $ 3.Aut(A_6) $ exists but the other two do not exist.
Connecting this back to the original question, the Schur cover of $ A_6 $ is $ 6.A_6 $ so the fact that the extension
$$
6.A_6.Out(A_6)=6.Aut(A_6)
$$
does not exist provides a counterexample to the conjecture in the original question.
Source: See "groups" sidebar or "standard generators" section at the top of this page https://brauer.maths.qmul.ac.uk/Atlas/v3/alt/A6/ to see which almost quasisimple groups are possible for $ A_6 $.
Best Answer
The infinite dihedral group is an example with property 1.