Can any arbitrary subset $ A⊂G $ have stabilzer ? If $ S $ be the Stabilizer of the $ A $ by $ G $. then I do know that
$$S = \{ s: sas^{-} = A \ ; \ ∀s∈G\ \,\ a∈A \} $$
Also since $S$ is a subgroup of $ G$, then $ |S|$ must be some multiple factor of $ |G|$. Now if we take arbitrary set $A$ such that $ |A| $ is not a factor of $|G|$, then $|A| ≠ |S|$. How can their order not be same and still we can have a Subgroup?
While $S$ will be a subgroup of $G$, it's not true that $|A| = |S|$, even if $A$ is a subgroup of $G$, even in the more general context of $G$ acting on any set, not just a subset of itself.
For a small example let $A = \{e\}$, or let $A = \{a\}$ for some $a \in G$ if you don't want $A$ to be a subgroup. In each case, if $G$ is abelian, then $S = G$, and certainly $|A| \neq |S|$.