Conjugate stabilisers Let G act on X and let x,y $\in$ X
I know that If x and y belong to the same orbit then $<x>=<y>$ then their stabilisers $G_x$ and $G_y$ are conjugate subgroups of G.
This is the proof I have. 
Let $y \in <x>$. Then $y=x \wedge g$ for some $g \in G$ and I try to show that $$G_y=g^{-1} G_x g$$
I then have $$g^{-1} G_x g \subseteq G_y$$ and let $h \in G$
Then $$y \wedge g^{-1} h g=((y \wedge g^{-1}) \wedge h) \wedge g$$
$$=(y \wedge h) \wedge g$$
$$=y$$
So $g^{-1} h g \in G$ hence $$g^{-1} G_x g \subseteq G_y$$
Now $Gy \subseteq g^{-1} G_x g$ and let $k=G_y$
Then $$k=g^{-1}(g k g^{-1})g$$ where $x \wedge gkg^{-1}=((x \wedge g) \wedge k) g^{-1}$
$$=( y \wedge k)\wedge g^{-1}=y \wedge g^{-1}$$ $$=x$$ so $gkg^{-1} \in G_x$ so $$G_y \subseteq g^{-1} G_x g$$
Why can we say from $G_y=g^{-1} G_x g$ that $g^{-1} G_x g \subseteq G_y$g ?
Does this proof make sense?
 A: $(1)$ You can always find $g$ such that $g \cdot a = b$
$(2)\ \textrm{Stab}(b) = \textrm{Stab}(g \cdot a) = g\textrm{Stab}(a)g^{-1}$ (look below why this is true!)
$(3)\ x \in \textrm{Stab}(g \cdot a) \iff x(ga) = ga \iff (g^{-1}xg)a = a \iff g^{-1} x g \in G_a \iff x = gg'g^{-1} \iff x \in g\textrm{Stab}(a)g^{-1}$.
A: The goal of the proof is two show two sets are equal. If we call them $A$ and $B$, we want to show every element in $A$ is in $B$ and vice versa. Well if every element of $A$ is in $B$, then $A \subset B$. Likewise, if everything in $B$ is in $A$ then $B \subset A$.  
Thus if you can show $A \subset B$ and $B \subset A$, then you have shown $A=B$. That is the proof.
A: Look, suppose that we know that $x$ is some conjugate of $y$, say $x = g^{-1}yg$. Note this also means $y = gxg^{-1}$.
Now, take any element, say $a \in G_y$. By definition, this means:
$a^{-1}ya = y$.
We look at what $g^{-1}ag$ does when acting on $x$ by conjugation:
$(g^{-1}ag)^{-1}x(g^{-1}ag) = (g^{-1}a^{-1}g)x(g^{-1}ag) = (g^{-1}a^{-1})(gxg^{-1})(ag)$
$= (g^{-1}a^{-1})y(ag) = g^{-1}(a^{-1}ya)g = g^{-1}yg = x$.
Evidently, then, $g^{-1}G_yg \subseteq G_x$.
The reverse inclusion is similar, start with $b \in G_x$, and consider the action of $gbg^{-1}$ on $y$:
$(gbg^{-1})^{-1}y(gbg^{-1}) = (gb^{-1}g^{-1})y(gbg^{-1}) = (gb^{-1})(g^{-1}yg)(bg^{-1})$
$= (gb^{-1})x(bg^{-1}) = g(b^{-1}xb)g^{-1} = gxg^{-1} = y$.
This shows that $b = g^{-1}(gbg^{-1})g \in g^{-1}G_yg$, so $G_x \subseteq g^{-1}G_yg$.
