I am studying an article on symmetric configurations and I found a statement about group theory that I can not understand.
Definition: Each orbit is associated a conjugacy class of subgroups of $G$, called the orbit type of the orbit.
Affirmation: if an orbit has type $(H)$ say, then at least one of the points $x$ of the orbit has isotropy subgroup $G_x = H$.
Apparently the assertion results from the fact that $G_y=gG_xg^{-1}$ since that $y=gx$, because the author proof before the affirmation.
My question is: associated means that exists a bijection between each orbit and the conjugation class of a subgroup? how can I proof the affirmation? I've try to use that $G_y=gG_xg^{-1}$, but without sucess.
Notations:
$G$ is a group;
$H$ is a subgroup of G;
$G_x=\{g\in G: gx=x\}$;
orbit of $x$ over $G$ is $G.x=\{gx: g\in G\}$
The group $G$ acts with the set $X$ and $x\in X$.
Given an orbit $A$, the associated conjugacy class consists of the stabilizers $G_x$ of all the elements $x\in A$. It is easy to prove that (1) stabilizers are subgroups of $G$ and (2) the stabilizers of elements of $A$ constitute exactly one conjugacy class of subgroups of $G$.