If two elements are in the same orbit, show the stablilizer groups of those elements are conjugate subgroups There was a proof of this answered before that I don't understand. Here it is   

Suppose $x$ and $y$ are in the same orbit, so $x=gy$ for some $g$. If $h$ stabilizes $y$, then $ghg^{-1}x=ghy=gy=x$. This gives an injective homomorphism from $Stab(y)$ to $Stab(x)$. You can verify that $h\mapsto g^{-1}hg$ is its inverse, so the two groups are conjugate.  

I am to show $Stab(x) = gStab(y)g^{-1}$ by showing $$Stab(x) \le gStab(y)g^{-1} ; gStab(y)g^{-1} \le Stab(x) $$
so in the above proof $h \in G$ acts on $y \in X$ by $hyh^{-1} =y$ so $h \in gStab(y)g^{-1}$
I can see that the equality is true but not why it gives a homomorphism. Any further explanation would be very helpful. 

For the proof you cite. They define a map $\varphi : \mbox{Stab}(y) \to g\mbox{Stab}(y)g^{-1}$. For $h \in \mbox{Stab}(y)$
$$
 \varphi(h) := ghg^{-1}
$$
i.e. conjugation by $g$, and this gives a homomomorphism which is injective (maybe you try to check this for yourself!). By definition $\varphi(h) \in g\mbox{Stab}(y)g^{-1}$ for $h \in \mbox{Stab}(y)$ and clearly this map is surjective. The computation $ghg^{-1}x = ghy = gy = x$ shows that $\varphi(h) \in \mbox{Stab}(x)$. So taken 
together we have
$$
 g\mbox{Stab}(y)g^{-1} = \mbox{image}(\varphi) \le \mbox{Stab}(x)
$$
and so you have your injection from $\mbox{Stab}(y)$ to $\mbox{Stab}(x)$. An analog computation with the map $\psi(h) := g^{-1}hg$ gives the other inclusion $g^{-1}\mbox{Stab}(x)g \le \mbox{Stab}(y)$. Now noting that $\varphi$ and $\psi$ are inverse to each other we see that this implies $\mbox{Stab}(x) \le g\mbox{Stab}(y)g^{-1}$ by applying the inverse map on both sides.
So we have two things up to now:
1) $\mbox{Stab}(x) = g\mbox{Stab}(y)g^{-1}$
2) $\mbox{Stab}(x)$ and $\mbox{Stab}(y)$ are isomorphic.
By looking at the specific isomorphism given we see that we can refine 2) by saying the are even conjugate (as conjugation is just a special form of isomorphism). Combining 1) and 2) we see that the stabilizers of the elements $x$ and $y$ in the same orbit are conjugate by the element $g$ that maps them onto each other. QED
But maybe let me add a shorter, or more condensed proof, which I find easier myself. First convince yourself that conjuation gives an isomorphism, then if $x = yg$ we have
$$ 
 h \in \mbox{Stab}(x) \Leftrightarrow (yg)h = (yg) \Leftrightarrow yghg^{-1} = y \Leftrightarrow ghg^{-1} \in \mbox{Stab}(y) \Leftrightarrow h \in g^{-1}\mbox{Stab}(y)g
$$
where in the last step we have rearranged the conjugation operation (using the inverse map). Now by the above equivalences we see instantly that $\mbox{Stab}(x) = g^{-1}\mbox{Stab}(y)g$.
