TITLE: Stabilizer Conjugation
QUESTION [5 upvotes]: This may be a straightforward question, but if I have a group $G$ acting on a set $A$, and two elements $a,b\in A$ belong to the same orbit, how do I show that their stabilizers are conjugate. 
So far I know that  $a=gb$ for some $g\in G$. Do I just need to show that $g$ times some element of the stabilizer of $b$ is equal to a stabilizer of $a$?

REPLY [2 votes]: It is $b \in O(a) \Leftrightarrow \exists \bar g \in G \mid b= \bar g \cdot a$. Then:
\begin{alignat}{1}
\operatorname{Stab}(b) &= \{g \in G \mid g \cdot b = b\} \\
&= \{g \in G \mid g \cdot (\bar g \cdot a) = \bar g \cdot a\} \\
&= \{g \in G \mid (g \bar g) \cdot a = \bar g \cdot a\} \\
&= \{g \in G \mid \bar g^{-1}\cdot((g \bar g) \cdot a) = \bar g^{-1}\cdot(\bar g \cdot a)\} \\
&= \{g \in G \mid (\bar g^{-1}g \bar g) \cdot a = a\} \\
\tag 1
\end{alignat}
Now, call $g':=\bar g^{-1}g \bar g$; then, $g=\bar gg'\bar g^{-1}$ and $(1)$ reads:
\begin{alignat}{1}
\operatorname{Stab}(b) &= \{\bar gg'\bar g^{-1} \in G \mid g' \cdot a = a\} \\
&= \{\bar gg'\bar g^{-1} \in G \mid g' \in \operatorname{Stab}(a)\} \\
&= \bar g \operatorname{Stab}(a)\bar g^{-1}
\end{alignat}