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Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a smooth injective function with $\operatorname{det}[f'(x)]\not=0 $ for all $x\in\mathbb{R}^n$. Moreover assume that $f$ has no fixed points. Can $f$ have a periodic points i.e. a point x such that $f^k(x):=(f\circ f\circ\ldots\circ f)(x)=x$? The answer for $n=1$ is NO, but I don't see the answer for $n>2$.

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Let $h\colon \mathbb R\to\mathbb R$ be a smooth function with $h(0)\ne0$, $h(-x)=h(x)$ and $h(x)=0$ for $x>1$, for example $$ h(x)=\begin{cases}e^{-\frac1{1-x^2}}&\text{if }|x|<1\\ 0&\text{if }|x|>1\end{cases}$$ Then in $\mathbb R^2$ $$ f(x,y)=(x+h(y),-y)$$ has no fixed points, but any point with $|y|>1$ has period $2$ (and the properties smooth, injective, $\det f'\ne0$ are immediately verified).

In fact even $f(x,y)=(x+1-y^2,-y)$ is an example with periodic points but no fixed points

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