The thought came from the following problem:
Let $V$ be a Euclidean space. Let $T$ be an inner product on $V$. Let $f$ be a linear transformation $f:V \to V$ such that $T(x,f(y))=T(f(x),y)$ for $x,y\in V$. Let $v_1,\dots,v_n$ be an orthonormal basis, and let $A=(a_{ij})$ be the matrix of $f$ with respect to this basis.
The goal here is to prove that the $A$ is symmetric. I can prove this easily enough by saying:
Since $T$ is an inner product, $T(v_i,v_j)=\delta_{ij}$.
\begin{align*} T(A v_j,v_i)&=T(\sum_{k=1}^n a_{kj} v_k,v_i)\\ &=T(a_{1j} v_1,v_i) + \dots + T(a_{nj} v_n,v_i)\\ &=a_{1j} T(v_1,v_i) + \dots + a_{nj} T(v_n,v_i)\tag{bilinearity}\\ &=a_{ij}\tag{$T(v_i,v_j)=\delta_{ij}$}\\ \end{align*}
By the same logic,
\begin{align*} T(A v_j,v_i)&=T(v_j,A v_i)\\ &=T(v_j,\sum_{k=1}^n a_{ki} v_k)\\ &=T(v_j,a_{1i} v_1)+\dots+T(v_j,a_{ni} v_n)\\ &=a_{1i} T(v_j,v_1)+\dots+a_{ni} T(v_j,v_n)\\ &= a_{ji}\\ \end{align*}
By hypothesis, $T(A v_j,v_i)=T(v_j,A v_i)$, therefore $a_{ij}=T(A v_j,v_i)=T(v_j,T v_i)=a_{ji}$.
I had this other idea though, that since $T$ is an inner product, its matrix is positive definite.
$T(x,f(y))=T(f(x),y)$ in matrix notation is $x^T T A y=(A x)^T T y$
\begin{align*} x^T T A y &= (A x)^T T y\\ &=x^T A^T T y\\ TA &= A^T T\\ (TA)^T &= (A^T T)^T\\ A^T T^T &= T^T A\\ TA &= A^T T^T\tag{T is symmetric}\\ &= (TA)^T\tag{transpose of matrix product}\\ \end{align*}
This is where I got stuck. We know that $T$ and $TA$ are both symmetric matrices. Clearly $T^{-1}$ is symmetric. If it can be shown that $T^{-1}$ and $AT$ commute, that would show it.