we know that Isometric deformation implies $^=$ but does the opposite works?

Question

Let $f: \mathbb{R}^m \rightarrow \mathbb{R}^n$ be an isometric embedding ($n \geq m$). Then according to the answer here : $J^T J=I_{n}$, where $J$ is the jacobian matrix of $f$, and $T$ refers to the transpose of a matrix.

I am wondering if the opposite also works, meaning:

if $J^T J=I_{n}$ then $f$ is an isometric embedding?