About actions of ${\rm Aut}(G)$, conjugacy classes and inner automorphisms of said group

The action of $\mathrm{Aut}(G)$ on $G$ is simply $\varphi\cdot g = \varphi(g)$.

The conjugacy class of $x\in G$ is $\mathcal{C}_x=\{gxg^{-1}\mid g\in G\}$. You want to verify that applying $\varphi$ to each element of $\mathcal{C}_x$ will result precisely in $\mathcal{C}_{\varphi(x)}$. You will need to use that $\varphi$ is a group homomorphism, and that it is invertible.

For the second part, start by showing that the set in question is a subgroup of $\mathrm{Aut}(G)$: the identity morphism lies in the set, and if $\varphi$ and $\psi$ both lie in the subset, then so does $\varphi\circ\psi^{-1}$. Just check directly what happens with the inner automorphisms.

Finally, show that if $\varphi$ lies in the subset, and $\sigma$ is any automorphism, then $\sigma^{-1}\varphi\sigma$ is in the subgroup; that means that you need to show that if $\mathcal{C}_x = \mathcal{C}_{\varphi(x)}$, then $\mathcal{C}_{\sigma(x)} = \mathcal{C}_{\varphi(\sigma(x))}$, using the first part.

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 $\varphi(\mathscr C_x)=\mathscr C_{\varphi(x)}$.