let $f:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ be $C^{1}$ function such that $\forall x \in \mathbb{R}^{n} $ $df_{x}$ is isometry and $\mathbb{R}^{n}$ is with his euclidean structure , then $f$ must be an isometry
by local inversion theorem $\forall x \in \mathbb{R}^{n}$ $\exists$ $V_{x} $ (neighborhood of x) such that $\forall$ $y$ $\in$ $V_{x} $ we have $\left \| f(y)-f(x) \right \| = \left \| y-x \right \|$: since $f:V_{x} \rightarrow f(V_{x})$ is bijective and $df_{x}^{-1}$ is also isometry then $\left \| f(y)-f(x) \right \| \leq \left \| y-x \right \|$ and $\left \| f^{-1}(y)-f^{-1}(x) \right \| \leq \left \| y-x \right \|$ so the conclusion follows .
but now we need this result $\forall y \in \mathbb{R}^{n}$ i think to preuve that the set A={$y \in\mathbb{R}^{n}\mid$$\left \| f(y)-f(x) \right \| = \left \| y-x \right \|$ } is open and closed set and since $\mathbb{R}^{n}$ is connected we wille have $\mathbb{R}^{n}=A$ but i don't know how i can do this
