Transitive group action implies conjugate stabilizers even when group is infinite? The same question has been asked. But answer doesn't address my concerns here.
This was an exercise in textbook. They didn't mention that the group order has to be finite. 
It is immediately obvious to me that take two stabilizers $G_x$, $G_y$, we may show that for the group element $g$ for which $gx=y$, $gG_xg^{-1} \subset G_y$. Furthermore, we can show $gG_x g^{-1}$ is isomorphic to $G_y$. But still that doesn't imply $gG_xg^{-1}=G_y$ (e.g. integer group $nZ$ and $2nZ$) unless $|G|<\infty$. Is this problem wrong in that $G$ must be finite? How would you prove the infinite case?

$gx=y \Rightarrow g^{-1}y=x \Rightarrow g^{-1}G_yg \le G_x \Rightarrow G_y \le gG_xg^{-1}$.
