QUESTION: Defining $X=\{hxh^{-1} : h \in G\}$ as a conjugate set. We say that a group $G$ acts over $X$ if
$$
f\colon\begin{array}[t]{ >{\displaystyle}r >{{}}c<{{}} >{\displaystyle}l }
G\times X &\to& X \\
(g, hxh^{-1}) &\mapsto& f(g, hxh^{-1})=ghxh^{-1}g^{-1}
\end{array}
$$
RIGHT ANSWER:
- $O_{G}(x)=X$ is the G-orbit of $x$;
- $G_{x}=\{g \in G: gxg^{-1}=x\}$ is the stabilizer of $x$ in $G$.
MY QUESTION:
How to find this answers? I didn't understand how to apply this two following definitions to find such answers:
Definition (Stabilizer): Let $G$ be a finite group such that acts over a set X. Given an $x \in X$ (written $G_{x}$) the stabilizer of $x$ is the set $$G_{x}=\{g \in G: gx=x\}.$$
Definition (G-Orbit): Let $G$ be a finite group such that acts over a set X. We definie $G-\text{orbit}$ of $x$ (written $O_{G}(x)$) as the set $$O_{G}(x) = \{gx: g \in G\}.$$
Best Answer
I think it's misleading to call $X$ with this symbol, because one might be led to think of $x$ as a generic element of $X$, which is not the case. Rather, for a given $x \in G$, call:
$$S_x:=\{hxh^{-1}, h \in G\}\subseteq G \tag 1$$
So, for a given $x \in G$, your action is a $G$-action on $S_x$ and, for a given $s \in S_x$, orbit's general definition states that:
$$O_G(s)=\{f(g,s), g\in G\} \tag 2$$
which in your case reads:
$$O_G(s)=\{(gh_s)x(gh_s)^{-1}, g\in G\} \tag 3$$
where $h_s$ is such that $s=h_sxh_s^{-1}$. Now, the map $\varphi\colon G\to G$, defined by $g \mapsto \varphi(g):=gh_s$, is (in particular) surjective, since $\forall g'\in G, g'=\varphi(g'h_s^{-1})$; therefore, as $g$ spans $G$, $g'=gh_s$ does so, and thence $(3)$ becomes:
$$O_G(s):=\{g'xg'^{-1}, g'\in G\}=S_x \tag 4$$
(whence the action is transitive.)
Moreover, again for a given $s \in S_x$, by definition of stabilizer we get:
\begin{alignat}{1} \operatorname{Stab}_G(s)&=\{g\in G\mid f(g,s)=s\} \\ &=\{g\in G\mid (gh_s)x(gh_s)^{-1}=h_sxh_s^{-1}\} \\ &=\{g\in G\mid (h_s^{-1}gh_s)x(gh_s)^{-1}h_s=x\} \\ &=\{g\in G\mid (h_s^{-1}gh_s)x(h_s^{-1}gh_s)^{-1}=x\} \\ &=\{h_sg'h_s^{-1}\in G\mid g'xg'^{-1}=x\} \\ &=h_s\{g'\in G\mid g'xg'^{-1}=x\}h_s^{-1} \\ &=h_sC_G(x)h_s^{-1} \\ \tag 5 \end{alignat}
where $C_G(x)$ is the centralizer of $x$ in $G$, and finally ($s=x \Rightarrow h_s \in C_G(x)$):
$$\operatorname{Stab}_G(x)=C_G(x) \tag 6$$