On the determination of a quadratic form from its isotropy group  Let $F:\mathbf{R}^n\rightarrow\mathbf{R}$ be a non-degenerate quadratic forms. Let
$$
O(F):=\{g\in GL_n(\mathbf{R}):F(gv)=F(v),\forall v\in \mathbf{R}^n\}
$$
be the isotropy group of $F$.
Q: So how does one prove in the simplest way possible that if $O(F)=O(G)$ then
there exists $\lambda\in\mathbf{R}^{\times}$ such that $F=\lambda G$?
P.S. I would like to have a proof that could be explained to undergraduate students who take an advanced linear algebra class.
 A: Without changing anything important, we can orthogonally diagonalize the Gram matrix of the first form, call that $A.$ Call the other matrix $C.$ Your hypotheses are that $A$ has nonzero elements on the diagonal, and for any invertible matrix $P$, we know
$$  P^T A P = A \; \Leftrightarrow \; P^T C P = P.  $$
Take two indices $i < j,$ with the square block at entries $ii, ij, ji,jj$ being
$$ A_2 \; = \;  
 \left(  \begin{array}{cc}
  a & 0  \\\
   0  & b  
\end{array} 
  \right)  ,
  $$ 
where the first case we do is $a,b > 0.$ Then we may take $P$ to have all 0 entries except at the same four positions,
 $$ P_2 \; = \;  
 \left(  \begin{array}{cc}
  \cos t & \sqrt{\frac{b}{a}} \sin t  \\\
  - \sqrt{\frac{a}{b}} \sin t  & \cos t  
\end{array} 
  \right)  ,
  $$ 
giving us $P^T A P = A.  $ You need to show the students that $P^T C P $ also alters in exactly the way we expect in those four positions. Then, with
$$ C_2 \; = \;  
 \left(  \begin{array}{cc}
  r & s  \\\
   s  & t  
\end{array} 
  \right)  ,
  $$ 
and  $P^T C P = C,  $ we write out the three equations, resulting in $t \left( \frac{a}{b} \right) = r$ and $s=0.$
Similar with both $a,b < 0.$ 
Now suppose we chose two indices with diagonal elements of opposite sign,
$$ A_2 \; = \;  
 \left(  \begin{array}{cc}
  a & 0  \\\
   0  & -b  
\end{array} 
  \right)  ,
  $$ 
then take
 $$ P_2 \; = \;  
 \left(  \begin{array}{cc}
  \cosh t & \sqrt{\frac{b}{a}} \sinh t  \\\
   \sqrt{\frac{a}{b}} \sinh t  & \cosh t  
\end{array} 
  \right)  ,
  $$
and again $P^T C P = C  $ tells us that the 2 by 2 block of $C$ is daigonal with the same ratio of diagonal entries.
Do this for all the $(n^2 - n)/2$ pairs of indices. We have shown that $C$ is diagonal and all ratios of diagonal elements match the ratios for $A.$
 Since $C$ is also nondegenerate, all the diagonal entries of $C$ are nonzero and a constant times the same element in $A.$ 
A: A relatively easy proof also follows from using the reflection identity:  First, define the inner product associated to $F$, namely $v\ \cdot_F\ w = {\frac12}\bigl(F(v{+}w)-F(v)-F(w)\bigr)$, and then, for any $v$ with $F(v)\not=0$, define the reflection in $v$ by 
$$
\rho^F_v(w) = w - 2\ \frac{v\ \cdot_F\ w}{v\ \cdot_F\ v}\ v
$$
This map $\rho^F_v$ belongs to $O(F)$, as is easy to verify.  By your assumption, it belongs to $O(G)$ as well.  This works out to imply 
$$
\bigl(v\ \cdot_F\ v\bigr)\bigl(v\ \cdot_G\ w\bigr) = 
\bigl(v\ \cdot_G\ v\bigr)\bigl(v\ \cdot_F\ w\bigr).
$$
Now if you replace $v$ by $v+tw$ in this equation and compare $t$-linear terms, you'll get
$$
\bigl(v\ \cdot_F\ v\bigr)\bigl(w\ \cdot_G\ w\bigr) = 
\bigl(v\ \cdot_G\ v\bigr)\bigl(w\ \cdot_F\ w\bigr),
$$
i.e., $F(v)G(w) = F(w)G(v)$, from which the result is obvious.
A: If $v=0$, then we have $F(v)= G(v)=0$ Hence obviously we have $F(v)= \lambda. G(v)$ for some non zero real $\lambda$.  
Now if $v\neq 0$, then for any $g\in GL_n(\mathbb R)$, we have $g(v)\neq 0$ and $g^{-1}(v)\neq 0$ and hence $F(g^{-1}(v))\neq 0$ and $G(g^{-1}(v))\neq 0$, as $F$ and $G$ are non-degenerate.
If $g\in O(F)$ we have $g\in O(G)$.  We have $\forall v\in \mathbb R^n, v\neq 0$,
$$ F(g(v))= F(v)\text{ and } G(g(v))= g(v)$$
$$F(g(v)). G(v)= F(v). G(g(v))$$
$$F(g^{-1}(g(v))). G(g^{-1}(v))= F(g^{-1}(v)). G(g.g^{-1}(v))$$
$$F(v)= \frac{F(g^{-1}(v))}{G(g^{-1}(v))}.G(v)$$
Edited:   But it doesn't says that $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ is independent of $v$. So proof is incomplete.
