Let $V$ and $W$ be $n$ dimensional vector spaces on $\mathbb R$ with bilinear forms $\langle,\rangle_V:\, V\times V\longrightarrow\mathbb R$ and $\langle,\rangle_W:\, W\times W\longrightarrow\mathbb R$. Suppose a bijective map $f:\,V\longrightarrow W$ satisfies \begin{align} \big<f(u),f(v)\big>_W\,=\,\langle u,v\rangle_V,\ \forall\,u,v\in V.\tag1 \end{align} Moreover, suppose $V$ has an orthonormal basis with respect to $\langle,\rangle_V$, say, $\mathcal B=(e_1,\cdots,e_n)$.
Here, $\mathcal B$ is orthonormal basis means forall $i=1,\cdots,n$, $$\langle e_i,e_i\rangle_V\,=\,\pm1\ \text{ and } \ \langle e_i,e_j\rangle_V\,=\,0,\ j\ne i.$$
The Mazur–Ulam theorem states that if $\langle,\rangle_V,\,\langle,\rangle_W$ are inner products then $f$ must be linear, moreover its representation matrix is orthogonal.
My wonder is if this theorem is true for bilinear forms that not needed to be inner product, like considered in this discussion. Here my answer :
First, we have for any $u,v\in V$, \begin{align*} \langle u,v\rangle_V\,&=\,\left<\sum_{i=1}^n u_ie_i\,,\sum_{i=1}^n v_ie_i \right>_V \\ &=\,\sum_{i=1}^n\sum_{j=1}^n u_iv_j\big<e_i,e_j\big>_V \\ &=\,\begin{bmatrix} u_1 & \cdots & u_n \end{bmatrix}\begin{bmatrix} \big<e_1,e_1\big> & \cdots & \big<e_1,e_n\big> \newline \vdots & \ddots & \vdots \newline \big<e_n,e_1\big> & \cdots & \big<e_n,e_n\big> \end{bmatrix}\begin{bmatrix} \,v_1 \, \\ \vdots \\ v_n \end{bmatrix} \\ &=\,\big[u\big]_{\mathcal B}^\top\eta \big[v\big]_{\mathcal B}.\tag2 \end{align*} Next, since $f$ is bijective, $f(\mathcal B)$ is also orthonormal basis of $W$. Thus for any $v\in V$, there exists $t_1,\dots, t_n\in\mathbb R$ such that \begin{align} f(v)\,=\,t_1f(e_1)+\cdots+ t_n f(e_n)\tag3 \end{align} this implies $\forall\, i=1,\dots,n $ \begin{align} \big<f(v),f(e_i)\big>_W\,&=\,\big<t_1f(e_1)+\cdots+ t_n f(e_n),f(e_i)\big>_W \\ &=\,t_1\big<f(e_1),f(e_i)\big>_W+\cdots+t_n\big<f(e_n),f(e_i)\big>_W \\ &=\,t_1\big<e_1,e_i\big>_V+\cdots+t_n\big<e_n,e_i\big>_V \\ &=\,t_i\langle e_i,e_i\rangle_V. \tag4 \end{align} On the other hand, we have \begin{align} \big<f(v),f(e_i)\big>_W\,&=\,\langle v,e_i\rangle_V \\ &=\,\big<v_1e_1+\cdots+v_ne_n,e_i\big>_V \\ &=\,v_1\langle e_1,e_i\rangle_V+\cdots+v_n\langle e_n,e_i\rangle_V \\ &=\,v_i\langle e_i,e_i\rangle_V.\tag5 \end{align} From (4) and (5), we deduce $t_i=v_i$ for all $i=1,\dots,n$. Therefore \begin{align} f\big(v_1e_1+\cdots+v_ne_n \big)\,=\,v_1f(e_1)+\cdots v_nf(e_n). \end{align} Therefore, $f$ is linear. Moreover, if \begin{align} A\,=\,\begin{bmatrix} \big[f(e_1)\big]&\cdots&\big[f(e_n)\big] \end{bmatrix} \end{align} then $A^\top\eta A=\eta$.
Is my proof correct ? I hope to receive opinions from people. Thanks.