I'm interested in finding out on which sets $X$ it is possible to define an action of $G=\mathbb{F}_{p} ^\times$ or $S_n$.
For any $x \in X$ follows by the orbit-stabilizer theorem
$$|G/Stab_G(x)|=|Gx|=|X|$$
Hence $|X|$ divides $|G|=p-1$ and for example it is not possible to define such an action on a set with an odd number of elements.
However I don't know if/why in general it is possible to define a transitive action on a set which doesn't contradict the relation above. Could anyone enlighten me?
Best Answer
Using the relation you wrote yourself, you can clearly see that $G$ acts transitively on a set with $m$ elements if and only if $G$ has a subgroup of index $m$.
Since $\mathbb{F}_p^\times\simeq \mathbb{Z}/(p-1)\mathbb{Z}$ is cyclic, this is equivalent to $m$ dividing $p-1$.
For $G=S_n$, I have no idea if we know the possible indices of subgroups, but I would not be surprised if this were an open problem (there is a lot we do not know about the subgroups of $S_n$, like their number for instance).
Something that we can say, though, is that if $S_n$ acts transitively on a set with $m$ elements with $m\leqslant n$, then $m$ must be $1$ (trivial action), $2$ (action though the sign of the permutation) or $n$ (standard tautological action), except if $n=4$, in which case there is an exceptional additional $m=3$ allowed (given by the fact that $S_4$ has the $V_4$ subgroup of double transpositions, such that $S_4/V_4\simeq S_3$). There is also an exceptional phenomenon when $n=6$: there is a transitive action of $S_6$ on a set of $6$ elements which is not equivalent to the standard action (this corresponds to the exterior automorphism of $S_6$); this does not add new values of $m$ though.
When $m>n$ I have no idea what can be said.