Relation between two normal charts on a Riemannian manifolds?

Question

Let $M$ be a Riemannian manifold and $p\in M$ a point. Let $\{E_{i}\}$ be an orthonormal basis of $T_{p}M$ and let $E\colon \mathbb{R}^{n} \to T_{p}M$ be the isomorphism sending $(x_{1},\ldots,x_{n})\mapsto \sum_{i=1}^{n}x_{i}E_{i}$. On a normal neighborhood $U$ of $p$ in $M$ (that is, the image under the exponential map $\exp_{p}$ of a star shaped open neighborhood $V$ of $0\in T_{p}M$) we can define a coordinate chart by $\varphi =E^{-1}\circ \exp_{p}^{-1}\colon U\to V\subset \mathbb{R}^{n}$.

As is clear, the normal chart depends on choosing an orthonormal basis $\{E_{i}\}$ of $T_pM.$ So let $\{F_i\}$ be another orthonormal basis for $T_pM.$ Let $F:(y_{1},\ldots,y_{n})\mapsto \sum_{i=1}^{n}y_{i}F_{i},$. Let $\psi:= F^{-1}\circ \exp_{p}^{-1}\colon U\to V\subset\mathbb{R}^{n}$

My question is: what can we say about $\psi\circ\phi^{-1}:V\to V?$ We note that: $\psi\circ\phi^{-1}=F^{-1}\circ E.$ I feel that it's the restriction to $V$ of an orthogonal transformation of $\mathbb{R}^{n},$ i.e. restriction of an orthogonal matrix, but I'm having a difficult time proving it (if it's correct!). Clearly, the map $\psi\circ\phi^{-1}$ is a bijective linear map, since both $E,F$ are so. But what I'm wondering here is: does this preserve the inner product in $R^n?$

EDIT: I answered myself in the affirmative.

Answer 1

Choosing an orthonormal basis here is somewhat beside the point. What you should do is use an $n$-dimensional inner product space $W$ (which is of course isometric to $\mathbb R^n$ once you choose coordinates). Then consider a pair of isometries from $W$ to the tangent space, and see how the corresponding maps from $W$ to the manifold are related. Choosing an orthonormal basis should be the last step, and not in $T_p M$ but rather in $W$.

Answer 2

Answering my own question. The answer is yes. Below is the argument.

$E:\mathbb{R}^n\to T_p(M)$ is an inner product space isometry between $\mathbb{R}^n$ (canonical inner product) and $T_pM$ (inner product from the Riemannian metric), because $E$ preserves norms:

$$||E(x_1\dots x_n)||_{T_pM}^2=||\sum_{i=1}^{n}x_iE_i||_{T_pM}^2=\sum_{i=1}^{n}x_i^2= ||(x_1\dots x_n)||_{\mathbb{R}^n}^2$$

Note that the above (especially, the second equality) is possible precisely because we chose an orthonormal basis $\{E_i\}$. Otherwise, it wouldn't be an isometry.

Similarly, $F:\mathbb{R}^n\to T_p(M)$ is an inner product space isometry between $\mathbb{R}^n$ and $T_pM.$

So $\psi\circ\phi^{-1}= F^{-1}\circ E$ is a linear (inner product preserving) isometry from $\mathbb{R}^n$ to itself, i.e. restriction of an orthogonal matrix.