Does the concept of “adjoint map” determine the metric up to scaling?

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 $$h=0$$.

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.