Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$ Consider the group $G = S_n$ acting on $X = \left\{1,2,...n \right\}$.  Let $x=n$.  How many elements does the stabilizer $\operatorname{Stab_G(x)}$ have?  To what group is this stabilizer isomorphic?
Attempt:  There is $(n-1)!$  elements in $\operatorname{Stab_G(x)}$.  There exists an isomorphism $\phi$ such that $\phi : \operatorname{Stab_G(x)} \rightarrow S_{n-1}$, however, what is the simplest way to prove this?  Clearly, the sets have the same size so that is not an issue (in terms of bijectivity) and I have checked some cases, but I can't see how to prove it in general, which I think would be equivalent to finding an explicit homomorphism $\operatorname{Stab_G(x)} \rightarrow S_{n-1}$
Is it correct to say that $\operatorname{Stab_G(x)} \leq G$ acts trivially on the set $X$?  If so, I thought about considering the bijection $f_g : X \rightarrow X$, given by $f_g(x) = g \cdot x = x,\,\, g \in \operatorname{Stab_G(x)}$ which I think is a homomorphism from $\operatorname{Stab_G(x)}$ to some symmetric group (since this group is finite and $f_g \in \operatorname{bij}(X) = S_{|X|}),$ but I am unsure
Many thanks.
 A: Maybe it's a good idea to think permutations as bijections of finite sets. In other words, $S_n$ is the group of bijections of the set $\{1,\ldots,n\}$ with composition as operation.
Let $\eta\in Stab_G(x)\subseteq S_n$ be such a bijection. Note that 
the map 
$$
\begin{array}{rccc}
\Phi(\eta):&\{1,\ldots,n-1\}&\longrightarrow&\{1,\ldots,n-1\}\\
 &a&\longmapsto&\eta(a)
\end{array}
$$
is a bijection and therefore $\Phi(\eta)\in S_{n-1}$. Then $\Phi:Stab_G(x)\rightarrow S_{n-1}$ is the isomorphism you are looking for. To see that it indeed is an isomorphism you can try to construct its inverse, which should send
$\psi$ to the map $\widetilde{\psi}$ given by $\widetilde{\psi}(a)=\psi(a)$ if $a\in\{1,\ldots,n-1\}$ and $\widetilde{\psi}(a)=n$ if $a=n$.
A: Or; you can use the Orbit-Stablizer Equation. Moerover to @A. Bellmunt's way, when you want to show two permutation groups are isomorphic, you have to consider additional points there. See 1.
A: If you write the elements that fix $n$ in cycle notation, you will notice that these are precisely the elements that do not contain the symbol $n$ in any of your cycles (when simplified of course).  
So the isomorphism should be clear.  
A: For every $x\in X:=\{1,\dots,n\}$, set $Y:=X\setminus\{x\}$. Then, $\operatorname{Stab}(x)\stackrel{\varphi}{\cong} S_Y$ via $\sigma\mapsto\varphi(\sigma):=\sigma_{|Y}$. In fact:

*

*injectivity: $\varphi(\sigma)=\varphi(\tau)\Longrightarrow \sigma_{|Y}=\tau_{|Y}\stackrel{\sigma(x)=x=\tau(x)}{\Longrightarrow}\sigma=\tau$;

*surjectivity: for any given $f\in S_Y$, define $\sigma\in S_n$ by $(\sigma(x):=x) \wedge (\sigma_{|Y}:=f)$; therefore, $\sigma\in\operatorname{Stab}(x)$ and $\varphi(\sigma)=f$;

*operation-preserving: for $y\in Y$:
\begin{alignat}{1}
(\sigma\tau)(y) &= \sigma(\tau(y)) \\
&\stackrel{y\in Y}{=}\sigma(\tau_{|Y}(y)) \\
&\stackrel{\tau_{|Y}(y)\in Y}{=}\sigma_{|Y}(\tau_{|Y}(y)) \\
&=(\sigma_{|Y}\tau_{|Y})(y) \\
\end{alignat}
On the other hand, $y\in Y\Longrightarrow (\sigma\tau)(y)=(\sigma\tau)_{|Y}(y)$. Therefore:
$$\varphi(\sigma\tau)=(\sigma\tau)_{|Y}=\sigma_{|Y}\tau_{|Y}=\varphi(\sigma)\varphi(\tau)$$
Now, note that $|Y|=n-1$, and hence $S_Y\stackrel{\phi}{\cong} S_{n-1}$ for $\phi(f):=\mathscr gf\mathscr g^{-1}$, where $\mathscr g\colon Y\longrightarrow \{1,\dots,n-1\}$ is any bijection. So, finally, for every $x\in X$:
$$\operatorname{Stab}(x)\stackrel{\phi\varphi}{\cong} S_{n-1}$$
