Let $f(x)=A x+b$ be an isometry $\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$, where $A$ is an $n \times n$ matrix and $b \in \mathbb{R}^{n}$. What are the possible values of $\operatorname{det} A$ ?

Let $I$ denote the $n \times n$ identity matrix. Show that if $n=2$ and $\operatorname{det} A>0$, but $A \neq I$, then $f$ has a fixed point. Must $f$ have a fixed point if $n=3$ and $\operatorname{det} A>0$, but $A \neq I ?$ Justify your answer.