Isometry of translation (does $x+v \mapsto y+v$)

Question

Consider $f:\mathbb R^n\rightarrow \mathbb R^n$ where $f\in E_n$ and $E_n$ is the group of isometries of $\mathbb R^n$. Given $x,v\in \mathbb R^n$ arbitrary vectors of $\mathbb R^n$ and $y=f(x)$.
Does $f(x+v)=y+v$? If so, why?

Answer 1

Not necessarily. Take$$\begin{array}{rccc}f\colon&\mathbb R^2&\longrightarrow&\mathbb R^2\\&(x,y)&\mapsto&(-y,x).\end{array}$$Then $f$ is an isometry. But, if you take $x=(1,0)$, $y=(0,1)$, and $v=(1,1)$, then $y=f(x)$, but $f(x+v)\neq y+v$.

Answer 2

Not necessarily. But a similar statement is true:

Every isometry of $\mathbb{R}^n$ $f$ can be written in a form $f(x)=Mx+a$ where $M\in\mathbb{R}^{n\times n}$ is an orthogonal matrix and $a\in\mathbb{R}^n$ is a vector. Plugging in $x+v$ we get:

$$f(x+v)=M(x+v)+a=f(x)+Mv$$