Let $(M, g)$ be a Riemannian Manifold. Let $U \subseteq M$ and $V \subseteq \mathbb{R}^n$ be open sets, and let $\phi: U \rightarrow V$ be a diffeomorphism. Additionally, let $x^i: U \rightarrow \mathbb{R}$ be the coordinate maps such that $\phi = (x^1, ..., x^n)$. Define the top form $\omega := dx^1 \wedge \cdots \wedge dx^n$ to be a volume form on $U$ if, for any isometric open sets $A, B \subseteq U$, we have \begin{align*} \int_A \omega = \int_B \omega. \end{align*} This is not a definition I saw anywhere, I just came up with it as it seems natural.
Here is my question: Can we prove that $\frac{\partial}{\partial x^1}, ..., \frac{\partial}{\partial x^n}$ is an orthonormal basis for the tangent space at each point? This seems true intuitively. Equivalently: can we prove that $g$ is the identity matrix?
If $J$ is the Jacobian matrix of an isometry $f$, then $\det(J) = 1$ and \begin{align*} g_{f(p)} = J^T g_p J \end{align*} for all $p$. Thus, $\det(g)$ must be a constant map $U \rightarrow \mathbb{R}$. This feels like a step in the right direction, but doesn't rule out the possibility that $g$ is not the identity matrix.
This is all just out of curiosity.
No, defining $y_1 = x_1 + x_2$ and $y_i = x_i$ for $i > 1$ yields an identical top-dimensional form $$\mathrm{d}y_1 \wedge \cdots \wedge \mathrm{d}y_n = \mathrm{d}x_1 \wedge \cdots \wedge \mathrm{d}x_n + \mathrm{d}x_2 \wedge \mathrm{d}x_2 \wedge \cdots \wedge \mathrm{d}x_n = \omega.$$
If $\frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_n}$ is orthonormal, then $\frac{\partial}{\partial y_1}, \ldots, \frac{\partial}{\partial y_n}$ isn't, since
$$\left\langle\frac{\partial}{\partial y_1}, \frac{\partial}{\partial y_2}\right\rangle = \left\langle\frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2} - \frac{\partial}{\partial x_1}\right\rangle = -1.$$