Once I have managed to prove that every function $f:\mathbb{R}^n\to\mathbb{R}^n$ such $f(0)=0$ and $|f(u)-f(v)|=|u-v|$ for every $u,v\in\mathbb{R}^n$ is a linear operator and then orthogonal. Which I could conclude that every function that preserves distance in $\mathbb{R}^n$ has the form $g(v)=Av+b$, where $A:\mathbb{R}^n\to\mathbb{R}^n$ is an orthogonal (linear) operator and $b\in\mathbb{R}^n$ is a constant vector.
But I saw this question
Let $E$ be a vector space with inner product. Prove that every isometric immersion (preserves distance) $f:\mathbb{R}\to E$ has the form $f(t)=a+t\cdot u$ where $a,u\in E$ and $|u|=1$.
The relation is clear, but I don't know how to prove that one.
First, note that if $f$ is an isometric immersion, its derivative has unit length: $$ |f'(t)| = \lim_{\delta t\to0}\left| \frac{f(t+\delta t) - f(t)}{\delta t}\right|=1. $$ If we can show that it also has constant direction then we can write $f(t) = f(0) +t. u$ where $u = f'(t)$ for any $t$.
For every three points $t_1<t_2<t_3$, by triangle inequality we have $$ t_3-t_1 = |f(t_3) - f(t_1)| \leq |f(t_3) - f(t_2)|+|f(t_2) - f(t_1)|=t_3-t_2+t_2-t_1 = t_3-t_1 $$ However, the equality can only be satisfied if $f(t_3) - f(t_2)$ and $f(t_2) - f(t_1)$ are parallel (see this post as mentioned by Lucio Tanzini in the comments). If that is true for every $t_1<t_2<t_3$, then the direction of $f'$ has to be constant.