Suppose we have a group $G$ acting of a set $X$, and take $x \in X$. It is well known that any conjugate of the stabilizer $G_x$ is iself a point stabilizer, specifically for $g \in G$, we have $gG_xg^{-1} = G_{gx}$.
However I was wondering, what happens if we take an automorphism $\phi \in \operatorname{Aut}(G)$ which is not necessarily inner? Is then $\phi (G_x)$ also a point stabilizer?
By the way I am asking this for finite groups, so even if there is a proof of this which uses finiteness, it is more than welcome.
No. Consider the action of the Klein $4$-group $C_2\times C_2$ on a two element set $\{x,y\}$ given by letting every element of the form $(0,\Box)$ fix both $x$ and $y$, and every element of the form $(1,\Box)$ exchange $x$ and $y$; this is just the action of $S_2$ on $\{x,y\}$ inducing an action of $C_2\times C_2$ via the corresponding map $C_2\times C_2\to S_2$.
The point stabilizers of both $x$ and $y$ is $\{(0,0), (0,1)\}$. Any other subgroup of order $2$ is the image of this point stabilizer under an automorphism of $C_2\times C_2$. So there are automorphisms of $C_2\times C_2$ that map $G_x$ to a subgroup that is not a point stabilizer.