Inner and Outer Group Automorphisms I have a question about something I believe the naming convention for group automorphisms suggests. From my understanding the inner automorphisms defined on a group $G$ are those automorphisms $\varphi: G \to G$ where there is some $g \in G$ where $\varphi$ is equivalent to the action of $g$ on $G$ by conjugation.
Those automorphisms which are not inner automorphisms ($\text{Inn }G$), are outer automorphisms. To me, this suggests as if (informally) there is a larger group $A$ where $G \trianglelefteq A$, and those "outer" automorphisms are simply equivalent to the action of some element $a \in A/G$ on $G$ by conjugation (It's called outer because $a$ is outside of $G$). Note that I'm stating $G$ is a normal subgroup of $A$ since we want $G$ to be normalized by all elements of $A$ so that the action of $a$ by conjugation limited to $G$ becomes an automorphism.
This idea is not discussed in the textbook I am studying. So my question is if the following theorem is valid, and how one goes around proving it.
Proposition: For every group $G$ there exists groups $G'$ and $A'$ such that $G \cong G'$ and $G' \trianglelefteq A'$ where the following condition holds:
$$ \text{Aut }G \cong \text{Inn}\ A' $$

Here  are some further finite counterexamples. One such is an extra-special group $G$ of order $p^3$ and exponent $p$ for an odd prime $p$.
Then ${\rm Aut}(G) = p^2\!:\!{\rm GL}(2,p) = {\rm AGL}(2,p)$ is a split extension of an elementary abelian group of order $p^2$ by ${\rm GL}(2,p)$ with the natural induced action.
Here is sketch proof that no group $A$ exists with $G \unlhd A$ and ${\rm Aut}(G) \cong {\rm Inn}(A) = A/Z(A)$. I can give more details if requested.
If $Z(G) \not\le Z(A)$, then $G \cap Z(A) = 1$, so $A/Z(A)$ has a normal subgroup isomorphic to $G$. But ${\rm Aut}(G)$ has no such subgroup.
So $Z(G) \le Z(A)$ and in fact $Z(A) \cap G = Z(G)$, and $G/Z(A)$ is the unique elementary abelian normal subgroup of $A/Z(A)$ of order $p^2$. But elements of $A$ that map onto elements of ${\rm GL}(2,p)$ with determinant not equal to $1$ induce nontrivial actions on $Z(G)$, which contradicts $Z(G) \le Z(A)$.
