Let $G$ be a group of order $27$ that acts on a set $X$ with $135$ elements. Prove that for each $x_1,x_2 \in X$ ${\rm Stab}_G(x_1) ={\rm Stab}_G(x_2)$.
I really can't think of anything useful here. I think it suffices to prove that every element $g \in G$ either fixes nothing or all the elements of $S$. I can't seem to go much further than that though. No need for the full solution, just a hint will do.
Thanks.
As an 'answer' to this, let's answer the question 'if $|G|=n$ and $|X|=m$, when are we guaranteed that all stabilizers of elements of $X$ have the same order?' Equivalently, all orbits should have the same length, which makes it easier to understand.
Let $p$ be the smallest prime dividing $|G|$. If $|X|<p$ then all orbits must be of length $1$, and if $|X|=p$ then either all orbits have length $1$ or there is a single orbit of length $p$. So there is a positive answer if $|X|\leq p$.
If $|X|\geq p+1$ then we may take $G$ cyclic of order $n$, whence it has an action on $p$ points. Let $G$ act with a single orbit of length $p$, and $|X|-p$ orbits of length $1$.
If one requires two distinct orbits of length greater than $1$, then one may suitably modify this with $|G|=pqm$ where $q$ is the second smallest prime dividing $|G|$, or $p^2$ if $p^2\mid |G|$ and $p^2$ is smaller than the second smallest prime. Now the two smallest orbits have size $p$ and $q$, so the result holds if $|X|<p+q$.
(Of course, this is the statement for all groups of order $|G|$. If we restrict to specific groups then the answer will vary, and it depends on the minimal degree of a permutation representation.)