Defining $X=\{hxh^{-1} : h \in G\}$ as a conjugate set. We say that a group $G$ acts over $X$ if $$
f\colon\begin{array}[t]{ >{\displaystyle}r >{{}}c<{{}}  >{\displaystyle}l } 
          G\times X &\to& X \\ 
          (g, hxh^{-1}) &\mapsto& f(g, hxh^{-1})=ghxh^{-1}g^{-1} 
\end{array}
$$
Prove that  this action is transitive.
My answer: An action is transitive when there is $x \in X$ such that the $G-\text{orbit}$ of x in G, that is, $O_{G}(x)=\{(gh)x(gh)^{-1}:g, h \in G\}= X = \{hxh^{-1}, h \in G\}$.
But I am not entirely convinced that my answer is right ....
How do I disappear with the $g$ on the right and on the left in the $G-\text{orbit}$ of x in G set?
Only then, I will make sure that these sets are really the same.

Suppose $x'=gxg^{-1}\in X$. Then this is equal to $f(g,x)$. Since this describes all elements of $X$, we are done: given any $x'\in X$, there exists some $g\in G$ such that $f(g,x)=x'$.
If you only know that $hxh^{-1}\in X$ for some $h$, note that $f(h^{-1},hxh^{-1})=x$, so $x\in X$ and we can use the argument above. Combining, we have that
$$f(gh^{-1},hxh^{-1})=gxg^{-1}$$
