Show that $ \text{Stab}\left(v\right)\approx O_{n-1}\left(\mathbb{R}\right) $ for $v \in \mathbb{R}^n$

Consider the natural operation of the group of orthogonal matrices on $\mathbb{R}^n$

$$ O_{n}\left(\mathbb{R}\right)=\left\{ A\in GL_{n}\left(\mathbb{R}\right)|AA^{t}=I\right\} $$ And the operation $ \varphi:O_{n}\left(\mathbb{R}\right)\to\text{Sym}\left(\mathbb{R}\right) $ is given by $$ \varphi\left(A\right)\left(v\right)=Av $$

I want to prove that for any $ 0\neq v\in\mathbb{R}^{n} $ we have $ \text{Stab}\left(v\right)\approx O_{n-1}\left(\mathbb{R}\right) $

Where $ \text{Stab}\left(v\right)=\left\{ A\in O_{n}\left(\mathbb{R}\right):Av=v\right\} $.

Here's what I think:

Fix $ v\in \mathbb{R} $ and fix an orthonormal basis $ B=\left(\frac{v_{1}}{||v_{1}||},v_{2},v_{3},...,v_{n}\right) $.

Now for any $A\in \text{Stab}\left(v\right) $, The image of B would be an orthonormal basis of the form $ C=\left(\frac{v_{1}}{||v_{1}||},u_{2},...,u_{n}\right) $.

Define a linear transformation by $$ \begin{cases} T\left(v_{2}\right)=u_{2}\\ T\left(v_{3}\right)=u_{3}\\ \vdots\\ T\left(v_{n}\right)=u_{n} \end{cases} $$

Now $ [T]_{C}^{B} $ is $I_{n-1}$ and my intuition is that $ [T]_{E}^{E} $ where $ E $ is the standard basis of $\mathbb{R}^{n-1}$ would be the orthogonal matrix that I want to map $ A $ to.

Im not sure how to write a formal proof for this. I'll be very greatful for an idea.

Thanks in advance.

If $A\in\operatorname{Stab}(v)$ and $w\in v^\perp$, then, since $A\in O_n(\Bbb R)$, $Aw\in v^\perp$. So, you have a natural bijection from $\operatorname{Stab}(v)$ onto $\{A\in O_n(\Bbb R)\mid A.v^\perp\subset v^\perp\}$. And this space is homeomorphic to $O_{n-1}(\Bbb R)$; just fix an orthonormal basis $B=\{v_1,v_2,\ldots,v_{n-1}\}$ of $v^\perp$ and, if $A\in O_n(\Bbb R)$ is such that $A.v^\perp\subset v^\perp$, then $A|_{v^\perp}$ is a linear map whose matrix with respect to $B$ as an element of $O_{n-1}(\Bbb R)$.