Let $V$ be a real finite-dimensional vector space, and $g$ and inner product on $V$. $g$ induces a concept of "adjoint map" , i.e. a linear map $\text{Hom}(V,V) \to \text{Hom}(V,V)$ given by $S \to S^T$, where $S^T$ is defined by requiring
$$ g(Sx,y)=g(x,S^Ty).$$
Now, suppose that $h:V \times V \to \mathbb{R}$ is a symmetric bilinear map on $V$, satisfying $$ h(Sx,y)=h(x,S^Ty), \tag{1}$$ for every $S \in \text{Hom}(V,V)$ and $x,y \in V$.
Is it true that $h=\lambda g$ for some $\lambda \in \mathbb{R}$?
We can think of $g$ as an isomorphism $g:V \to V^*$. In that case $S^T:V \to V$ is obtained from the dual map $S:V^* \to V^*$, via $S^T=g^{-1}\circ S^* \circ g$. If $h$ is also non-degenerate, then condition $(1)$ implies that the "transpose map" w.r.t $h$ is the same as w.r.t $g$, i.e. $$ h^{-1}\circ S^* \circ h=S^T_h =S^T_g=g^{-1}\circ S^* \circ g,$$
i.e. conjugations by $h,g$ are the same. Does that forces $h=\lambda g$?
I do not assume that $h$ is positive or non-degenerate. (In particular, I am allowing $h=0$). However, we can prove that if $h$ is non-zero then it is non-degenerate: Suppose that $h(x,y)=0$ for every $y$. Then, $0=h(x,S^Ty)=h(Sx,y)$ for every $y$. If $x \neq 0$ then $Sx$ can be chosen to an arbitrary vector in $V$, which forces $h=0$.
The answer is positive. As mentioned in the question, an equivalent reformulation of the question is the following:
If $A^{-1}XA=B^{-1}XB$ for every $X$ ($A,B,X$ are real $n \times n$ matrices) then $A=\lambda B$. Indeed, we have $XAB^{-1}=AB^{-1}X$, so $C=AB^{-1}$ commutes with every matrix $X$, hence it must be a scalar multiple of the identity.