I've seen this question posed in various forms, but I've yet to come across a clear-cut, straightforward, satisfactory answer. There's something obvious I'm missing but it has frustratingly been eluding me.
Problem: Let the group $G$ act on the nonempty set $A$. If $a, b \in A$ are in the same orbit, i.e. $b = ga$ for some $g \in G$, then $G_b = gG_ag^{-1}$, i.e. the stabilizer of $b$ in $G$ is conjugate to the stabilizer of $A$ in $G$.
Since I'm trying to prove the equality of two sets, I want to show that each is a subset of the other.
For the reverse direction: Let $h$ be an element of $G_a$. Then $ghg^{-1}$ is an element of $gG_ag^{-1}$, and $$ghg^{-1}b = ghg^{-1}ga = gha = ga = b,$$ so $ghg^{-1}$ stabilizes $b$, i.e. $ghg^{-1} \in G_b$. Hence $ gG_ag^{-1} \subset G_b$.
Now for the forward direction: Let $y \in G_b$. Then $yb = b$. I should be able to show that $y = gxg^{-1}$, for some $x \in G_a$, which would prove that $G_b \subset gG_ag^{-1}$. But I cannot see how to do this.
$yb=b=yga=ga$ implies that $g^{-1}yga=a$ and $g^{-1}yg=h\in G_a$ and $y=ghg^{-1}, y\in gG_ag^{-1}$.