Let $G$ be a finite group and $A\subseteq G$ a subset. The left regular action of $G$ on itself induces a natural action on the powerset of $G$: $$G\times 2^G\rightarrow 2^G,(g,A)\longmapsto gA:=\{ga|a\in A\}. $$ The set stabilizer of $A$ in $G$ is defined via: $$G_A:=\{g\in G|gA=A\}. $$Note that $gA=A$ only has to hold setwise, not pointwise. (Thus, $G_A\neq\cap_{a\in A}G_a$.) Let us write $A^{-1}:=\{a^{-1}|a\in A\}$ and $|\cdot|$ for the number of elements. I conjecture, that in general
$$|G_A|\neq|G_{A^{-1}}|. $$
Do you know a (nice, maybe smallest) example for a finite group $G$ and a subset $A$ such that $|G_A|\neq|G_{A^{-1}}|$ (or equivalently: The $G$-orbits of $A$ and $A^{-1}$ have different sizes.)? Certainly, $G$ will have to be non-abelian. A̶l̶s̶o̶ ̶$̶|̶G̶|̶>̶2̶6̶$̶,̶ ̶a̶s̶ ̶c̶a̶l̶c̶u̶l̶a̶t̶i̶o̶n̶s̶ ̶w̶i̶t̶h̶ ̶G̶A̶P̶ ̶d̶e̶m̶o̶n̶s̶t̶r̶a̶t̶e̶.̶
Or: (if I'm mistaken) Can you prove that indeed $|G_A|=|G_{A^{-1}}|$?
Thank you.
Encouraged by anon, I am repeating the counter example from my comment as a proper answer to my own question:
An example showing that $|G_A|\neq|G_{A^{-1}}|$ is: $$ G:=A_4,A:=\{(),(1,2)(3,4),(1,2,3),(1,3,4)\}. $$(The set stabilizers of $A$ and $A^{-1}$ have sizes 1 and 2.) baharampuri has correctly pointed out a connection to the right action. The example demonstrates that for the action on $2^G$ stabilizer sizes (and thus orbit sizes) of one "point" $A\in 2^G$ may differ for left and right action.