Let $G$ be a group that acts on a set $X.$ Let $\alpha, \beta \in X$ be in the same orbit. If $G$ acts on $P(G)$ via conjugation, then $G_{\alpha}$ and $G_{\beta}$ are in the same orbit.
$P(G)$ is the power set of $G$ i.e the set of all subset of $G$
$G_{\alpha}$ = {$g\in G: \alpha .g=\alpha$}
In the same way, $G_{\beta}$ = {$g\in G: \beta .g=\beta$}
How can I prove this statement?
Recall that the orbit of $G_{\alpha}$ is equal to $\mathcal O_{G_{\alpha}} = \{g^{-1} h g \,|\, g \in G \text{ and } \alpha \cdot h = \alpha \}.$ To show $G_{\alpha}$ and $G_{\beta}$ are in the same orbit, then for some $ g \in G,$ we have that $G_{\alpha} \cdot g = G_{\beta}.$ Any hints?
If $\alpha$ and $\beta$ are in the same orbit of some action $\cdot$ , then $G_\alpha$ and $G_\beta$ (stabilizers) are conjugate in $G$. In fact, $\beta\in O(\alpha)\Rightarrow\exists \tilde g\in G\mid \beta=\tilde g\cdot \alpha$; but then, by action properties:
\begin{alignat}{1} G_\beta &= \{g\in G\mid g\cdot \beta=\beta\} \\ &= \{g\in G\mid g\cdot (\tilde g\cdot \alpha) =\tilde g\cdot \alpha\} \\ &= \{g\in G\mid (g\tilde g)\cdot \alpha =\tilde g\cdot \alpha\} \\ &= \{g\in G\mid \tilde g^{-1}\cdot((g\tilde g)\cdot \alpha) =\tilde g^{-1}\cdot(\tilde g\cdot \alpha\}) \\ &= \{g\in G\mid (\tilde g^{-1}(g\tilde g))\cdot \alpha =(\tilde g^{-1}\tilde g)\cdot \alpha\} \\ &= \{g\in G\mid (\tilde g^{-1}g\tilde g)\cdot \alpha =\alpha\} \\ &= \{\tilde gg'\tilde g^{-1}\in G\mid g'\cdot \alpha =\alpha\} \\ &= \tilde g\{g'\in G\mid g'\cdot \alpha =\alpha\}\tilde g^{-1} \\ &= \tilde gG_\alpha\tilde g^{-1} \\ \tag 1 \end{alignat}
So, $G_\alpha$ and $G_\beta$ are coniugate in $G$. This precisely means that they lie on the same orbit of the action of $G$ by conjugation on $\mathcal{P}(G)$, say $\star$ . In fact: $\tilde g\star G_\alpha=\tilde g G_\alpha\tilde g^{-1}\stackrel{(1)}{=}G_\beta$.