Proving that $\text{Stab}(y \cdot x \cdot y^{-1}) = y \cdot \text{Stab}(x) \cdot y^{-1}$ This question is taken from my exam for an introductory course to algebraic structures:

Let $G, \cdot$ be a group and let $x \in G$. We define the stabilizer
  of $x$ as $$ \text{Stab}(x) = \left\{ y \in G \mid y \cdot x \cdot
 y^{-1} = x \right\}. $$
(i) Give an example of a group $G$ and an element $x$ such that
   $\text{Stab}(x) \neq G. $
(ii) Prove that for every $x \in G, \text{Stab}(x)$ is a subgroup of
  $G$.
(iii) Prove that for all $x,y \in G$ we have $$\text{Stab}(y \cdot x
\cdot y^{-1}) = y \cdot \text{Stab}(x) \cdot y^{-1}$$

My attempt: I'd appreciate it if you could point out any mistakes. For shorthand, I will write $S(x) = \text{Stab}(x). $ 
(i) Take $G = \text{GL}_n(\mathbb{R})$, the general linear group of degree $n$. Take for $x$ an $n \times n$ matrix that is not the identity matrix. Then in general, it will not be the case that $S(x) = G$, since not all matrices commute with $x$.
(ii) Notice first that $S(x) \neq \emptyset$, since $e_{G} \in S(x). $ Now let $y \in S(x)$. Then $yxy^{-1} = x$, and so $xy^{-1} = y^{-1} x$ or $x = y^{-1}xy$. This shows that $y^{-1} \in S(x)$.  
Now let $y_1, y_2 \in S(x)$. Then $y_1 x y_1^{-1} = x$ and $y_2x y_2^{-1} = x$. From the second of these equations it follows that $y_1 y_2 x y_2^{-1} = y_1x$ or by multiplying on the right with $y_1^{-1}$, we have that $$ (y_1 y_2) x (y_2^{-1} y_1^{-1}) = y_1 x y^{-1} = x $$ where in the last step I used the first equation. Hence $(y_1 y_2) x (y_1 y_2)^{-1} = x$ and this proves that $y_1 \cdot y_2 \in S(x)$. So $S(x)$ is a subgroup of $G$.
(iii) This is the part I got stuck on. I want to prove first that $S(y \cdot x \cdot y^{-1}) \subset y \cdot S(x) \cdot y^{-1}. $ Let $y_1 \in S(yxy^{-1}). $ Then we have that $$ y_1 (yxy^{-1}) y_1^{-1} = yxy^{-1}. $$ But then I'm not really sure what to do, because I don't know how the elements from $y \cdot S(x) \cdot y^{-1}$ look like. If I have $y_1 \in y \cdot S(x) \cdot y^{-1}$, does that mean that $$ y(y_1 x y_1^{-1}) y^{-1} = yxy^{-1}$$ ? How do these elements look like?
 A: $w \in y \cdot S(x) \cdot y^{-1}$ means $w = y z y^{-1} $, where $z \in S(x)$.
Check $w \in S(y xy^{-1})$:
$w y xy^{-1} w^{-1} = yz y^{-1} y xy^{-1} {(yz  y^{-1})}^{-1} = yz xy^{-1} (yz^{-1}  y^{-1}) =  yz xy^{-1} yz^{-1}  y^{-1} =yz xz^{-1}  y^{-1}  = yxy^{-1}  $
The last equation is by $z\in S(x)$. 
So $w \in S(y xy^{-1})$.
And the proof is done.
Here is the proof about another direction:

 $ m\in S(y xy^{-1}) \implies my xy^{-1}m^{-1} = y xy^{-1} \implies x = y^{-1}m^{-1}y (x)y^{-1}my \implies (y^{-1}my)^{-1} x (y^{-1}my)$
 $\implies y^{-1}my \in S(x) \implies m \in y S(x)y^{-1}$

A: Note that:

$g \in y \cdot \text{Stab}(x) \cdot y^{-1}$ iff for some $h \in G$, we have that $g = yhy^{-1}$, where $hxh^{-1} = x$.




*

*$\boxed{\subseteq}:$ Choose any $g \in \text{Stab}(yxy^{-1})$ so that $g yxy^{-1}g^{-1} = yxy^{-1}$. Solving for $g$, notice that:
$$
g = y (xy^{-1}gyx^{-1}) y^{-1}
$$
Convenient! So let's take $h = xy^{-1}gyx^{-1}$. If $h \in \text{Stab}(x)$, then we're done. Indeed, notice that:
\begin{align*}
hxh^{-1}
&= (xy^{-1}gyx^{-1})x(xy^{-1}gyx^{-1})^{-1} \\
&= (xy^{-1}gy)(xy^{-1}g^{-1}yx^{-1}) \\
&= xy^{-1}(gyxy^{-1}g^{-1})yx^{-1} \\
&= xy^{-1}(yxy^{-1})yx^{-1} &\text{since } g \in \text{Stab}(yxy^{-1})\\
&= xxx^{-1} \\
&= x
\end{align*}
as desired.

*$\boxed{\supseteq}:$ Choose any $g \in y \cdot \text{Stab}(x) \cdot y^{-1}$ so that for some $h \in G$, we have that $g = yhy^{-1}$, where $hxh^{-1} = x$. If $g \in \text{Stab}(yxy^{-1})$, then we're done. Indeed, notice that:
\begin{align*}
gyxy^{-1}g^{-1}
&= (yhy^{-1})yxy^{-1}(yhy^{-1})^{-1} \\
&= (yhxy^{-1})(yh^{-1}y^{-1}) \\
&= y(hxh^{-1})y^{-1} \\
&= yxy^{-1} & \text{since } h \in \text{Stab}(x)
\end{align*}
as desired.
