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.

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.