As the action is on a set of classes (necessarily dealt with by means of representatives), we have first to prove that the map is well defined, namely that $x'\in \mathscr C_x \Rightarrow \varphi\cdot\mathscr C_{x'}=\varphi\cdot\mathscr C_{x}$. This in indeed the case, because:
\begin{alignat}{1}
\mathscr C_{\varphi(x')} &= \{g\varphi(x')g^{-1}, g\in G\} \\
&= \{g\varphi(g'xg'^{-1})g^{-1}, g\in G\} \\
&= \{g\varphi(g')\varphi(x)\varphi(g')^{-1}g^{-1}, g\in G\} \\
&= \{(g\varphi(g'))\varphi(x)(g\varphi(g'))^{-1}, g\in G\} \\
&= \{g''\varphi(x)g''^{-1}, g''\in G\} \\
&= \mathscr C_{\varphi(x)} \\
\end{alignat}
where the last but one equality follows from $g\mapsto g\varphi(g')$ being onto $G$, for every $g'\in G$. Next, $\varphi\cdot\mathscr C_x\in \mathscr C$ by definition. Furthermore, $Id_G\cdot \mathscr C_x=\mathscr C_{Id_G(x)}=\mathscr C_x$. Finally, $(\varphi\psi)\cdot\mathscr C_x=\mathscr C_{(\varphi\psi)(x)}=\mathscr C_{\varphi(\psi(x))}=\varphi\cdot(\psi\cdot\mathscr C_x)$. So, this is indeed a group action and $\mathscr F(G)=\operatorname{Stab}(\mathscr C_x)$ is a subgroup of $\operatorname{Aut}(G)$. Then:
\begin{alignat}{1}
\psi\operatorname{Stab}(\mathscr C_x)\psi^{-1} &= \{\psi\varphi\psi^{-1}\mid \varphi\cdot\mathscr C_x=\mathscr C_x\} \\
&= \{\rho\mid (\psi^{-1}\rho\psi)\cdot\mathscr C_x=\mathscr C_x\} \\
&= \{\rho\mid \rho\cdot (\psi\cdot\mathscr C_x)=\psi\cdot\mathscr C_x\} \\
&= \operatorname{Stab}(\psi\cdot\mathscr C_x)
\end{alignat}
so $\mathscr F(G)$ is not normal in $\operatorname{Aut}(G)$. Rather, here we get that the stabilizers of the points of one same orbit are all conjugate to each other. Last, if $\varphi\in\operatorname{Inn}(G)$, then $\exists g\in G$ such that $\varphi\cdot\mathscr C_x=\mathscr C_{\varphi(x)}=\mathscr C_{gxg^{-1}}=\mathscr C_x$, whence $\varphi\in\mathscr F(G)$. Therefore, $\operatorname{Inn}(G)\le\mathscr F(G)$. (Note that this holds for every conjugacy class considered in $\mathscr F(G)$, and hence $\operatorname{Inn}(G)\le\operatorname{ker}\phi$, where $\phi$ is the homomorphism $\operatorname{Aut}(G)\to S_\mathscr C$ equivalent to your action.)
Comment. If the action is defined by $\varphi\cdot\mathscr C_x:=\mathscr C_{\varphi(x)}$, then the proof of the good definition focuses on the invariance w.r.t. representative's choice, being the membership of $\varphi\cdot\mathscr C_x$ to $\mathscr C$ trivial (by definition); this is the above approach. Conversely, if the action is defined by $\varphi\cdot\mathscr C_x:=\varphi(\mathscr C_x)$, then the proof of the good definition focuses on the membership of $\varphi\cdot\mathscr C_x$ to $\mathscr C$, being the invariance w.r.t. to representative's choice trivial (as $x'\in\mathscr C_x\Rightarrow \mathscr C_{x'}=\mathscr C_x$). The effort in both approach is the same, and equal to proving that $\varphi(\mathscr C_x)=\mathscr C_{\varphi(x)}$.
Edit, 9/25/20: The suggestion I made at the end works.
Proposition: Let $G$ be a group of order $n$ (which may be infinite). Then $\text{Inn}(G)$ is precisely the kernel of the action of $\text{Aut}(G)$ acting on the set $\text{Hom}_{\text{HGrp}}(F_n, G)$ of (simultaneous) conjugacy classes of $n$-tuples of elements of $G$.
Proof. Suppose $\varphi \in \text{Aut}(G)$ acts trivially. Consider its action on the $n$-tuple given by every element of $G$. Fixing this $n$-tuple means fixing it up to conjugacy, which means there is some $g \in G$ such that $\varphi(h) = ghg^{-1}$ for all $h \in G$, which says precisely that $\varphi \in \text{Inn}(G)$. On the other hand, every element of $\text{Inn}(G)$ clearly acts trivially. $\Box$
Of course we can do much better than considering every element of $G$; it suffices to consider a generating set. But this construction is at least "canonical."
Here's an approach that maybe will seem like it doesn't tell you anything new but I'll extract something slightly more concrete out of it, which generalizes the suggestion to look at conjugacy classes. $\text{Out}(G)$ occurs naturally as the automorphism group of $G$ in a category we might call the homotopy category of groups $\text{HGrp}$. This category can be defined concretely as follows:
- objects are groups $G$, and
- morphisms $f : G \to H$ are conjugacy classes of homomorphisms, where two homomorphisms $f_1, f_2 : G \to H$ are identified (homotopic) iff there exists $h \in H$ such that $h f_1 = f_2 h$.
For example:
- $\text{Hom}_{\text{HGrp}}(\mathbb{Z}, G)$ is the set of conjugacy classes of $G$
- $\text{Hom}_{\text{HGrp}}(G, S_n)$ is the set of isomorphism classes of actions of $G$ on a set of size $n$
- $\text{Hom}_{\text{HGrp}}(G, GL_n(\mathbb{F}_q))$ is the set of isomorphism classes of actions of $G$ on $\mathbb{F}_q^n$
and so forth.
Now we can prove the more general fact that composition in this category is well-defined (that is, that the homotopy class of a composition of morphisms only depends on the homotopy class of each morphism), which implies in particular that the automorphism group $\text{Aut}_{\text{HGrp}}(G)$ of $G$ in this category is really a group, and of course this group is $\text{Out}(G)$.
So far this is just a slight extension and repackaging of the proof via conjugating by an inner automorphism, but the point is that this construction tells you what conjugating by an inner automorphism means. The homotopy category of groups has a second description, as follows:
- objects are Eilenberg-MacLane spaces $K(G, 1) \cong BG$, and
- morphisms $f : BG \to BH$ are homotopy classes of homotopy equivalences.
We get the ordinary category of groups if we instead insist that Eilenberg-MacLane spaces have basepoints and our morphisms and homotopies preserve basepoints. So the passing to conjugacy classes has to do with the extra freedom we get from throwing out basepoints. Here the incarnation of conjugacy classes $\text{Hom}(\mathbb{Z}, G)$ is the set of free homotopy classes of loops $S^1 \to BG$.
Anyway, all this suggests the following generalization of looking at conjugacy classes: we can look at the entire representable functor
$$\text{Hom}_{\text{HGrp}}(-, G) : \text{HGrp}^{op} \to \text{Set}.$$
By the Yoneda lemma, the automorphism group of this functor is precisely $\text{Aut}_{\text{HGrp}}(G) \cong \text{Out}(G)$. What this says is that an outer automorphism of $G$ is the same thing as a choice, for each group $H$, of an automorphism (of sets) of $\text{Hom}_{\text{HGrp}}(H, G)$, which is natural in $H$. We can furthermore hope that it's possible to restrict attention to a smaller collection of groups $H$; for example (and I haven't thought about this at all) maybe it's possible to restrict to the free groups $H = F_n$, which means looking at $\text{Hom}_{\text{HGrp}}(F_n, G)$, the set of conjugacy classes of $n$ elements of $G$ (under simultaneous conjugacy).
Best Answer
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:
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.