For the rest of this problem, we assume that $\varphi$ is a non-degenerate symmetric bilinear form on $E$, and we denote by $q$ its quadratic form.
Let $e = (e_1, \ldots, e_n)$ be a basis of $E$. We say that $e$ is $q$-orthogonal if and only if, for all $(i,j) \in \{1,\ldots,n\}^2$ with $i \neq j$, $\varphi(e_i, e_j) = 0$.
Suppose that $e$ is simultaneously $q$-orthogonal and $q'$-orthogonal. Show that, for all $i \in \{1,\ldots,n\}$, $e_i$ is an eigenvector of $h^{-1} \circ h'$.