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 $\forall v\in \mathbb R^n, v\neq 0$,

$$ F(g(v))= F(v)\text{ and } G(g(v))= g(v)$$ $$F(g(v)). G(v)= F(v). G(g(v))$$ $$F(g^{-1}(g(v))). G(g^{-1}(v))= F(g^{-1}(v)). G(g.g^{-1}(v))$$ $$F(v)= \frac{F(g^{-1}(v))}{G(g^{-1}(v))}.G(v)$$

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.

If $v=0$, then we have $F(v)= G(v)=0$ Hence obviously we have $F(v)= \lamda. G(v)$ for some non zero real $\lamda$.

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 $\forall v\in \mathbb R^n, v\neq 0$,

$$ F(g(v))= F(v)\text{ and } G(g(v))= g(v)$$ $$F(g(v)). G(v)= F(v). G(g(v))$$ $$F(g^{-1}(g(v))). G(g^{-1}(v))= F(g^{-1}(v)). G(g.g^{-1}(v))$$ $$F(v)= \frac{F(g^{-1}(v))}{G(g^{-1}(v))}.G(v)$$

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 $\forall v\in \mathbb R^n, v\neq 0$,

$$ F(g(v))= F(v)\text{ and } G(g(v))= g(v)$$ $$F(g(v)). G(v)= F(v). G(g(v))$$ $$F(g^{-1}(g(v))). G(g^{-1}(v))= F(g^{-1}(v)). G(g.g^{-1}(v))$$ $$F(v)= \frac{F(g^{-1}(v))}{G(g^{-1}(v))}.G(v)$$