RIGHT ANSWER:
First inclusion: if $a \in G_x$ then $gag^{-1} \cdot gx= gax=gx$ thus $gag^{-1} \in G_{gx}$.
Second inclusion: if $b\in G_{gx}$ then defining $a=g^{-1}bg$ we have $ax=g^{-1}bgx=g^{-1}gx=x$ therefore $a \in G_x$ so $b=gag^{-1} \in gG_{x}g^{-1}$.
MY ANSWER:
By hypotesis we have: left action of group $G$ on $X$, which means there is a function $f:G \times X \rightarrow X$ given by $(g, x) \longmapsto f(g, x):=gx$.
Now, let's consider these groups:
$G_x=\{h \in G: hx=x\}$
$G_{gx}=\{h \in G: hgx=gx\}$
$\color{red}{gG_xg^{-1}=\{ghg^{-1} \in G: ghg^{-1}x=x\}}$ (Did I write properly this group?)
For me, in my imagination what I need to proof is $G_{gx}=gG_xg^{-1}$ this is, $\{h \in G: hgx=gx\}=\{ghg^{-1} \in G: ghg^{-1}x=x\}$. So I started as follows:
$(\supseteq)$ Let be $h \in G_{gx}\implies hgx=gx\implies g^{-1}hgx=g^{-1}gx=x\implies g^{-1}hgx=x.$ Here we found that $g^{-1}hg=x \in G_x$. Then because $x=g^{-1}hgx$ thus (multipling by $g$ in the left side and by $g^{-1}$ in the right side) we have $gxg^{-1}=gg^{-1}hgxg^{-1}\implies gxg^{-1}=hgxg^{-1} \in gG_xg^{-1}$. (is my answer also right?)
$(\subseteq)$ Let be $ghg^{-1}\in gG_xg^{-1}$ then $ghg^{-1}x=x$. Now I don't know how to achieve that this element belongs to $G_{gx}$. Look, that I'm trying to take an element inside $gG_xg^{-1}$ and not inside $G_x$ as in the right answer given. Is it possible?
$h\in G_x\implies ghg^{-1}(gx)=gh(x)=gx\implies ghg^{-1}\in G_{gx}$.
You got hung up because you assumed $ghg^{-1}x=x$, which is incorrect.