Suppose $U \subseteq \mathbb{R}^m$ is some connected open subset and let $f: U \to U$ be a differentiable map with $f^n = \text{id}$ for some $n\in\mathbb{N}$. Let $\text{Fix}(f)=\{x\in U: f(x) = x\}$. Suppose that $\text{Fix}(f)$ contains some non-empty open set. Is it true, that $f$ has to be the identity map?
If $U$ is any topological space and $f$ only continuous this would be false in general. For example take a plus $+$ as topological space and reflect it across the horizontal axis. Under which conditions on $X$ would this be maybe true for topological spaces?
Based on Moishe Kohan's hints:
Assume f is smooth. Introduce a Riemannian metric $g:=\sum_{k=1}^n(f^k)^*h$, where $h$ is some metric on $U$. Then $f^∗g=g$, because $f^n=\text{id}$, so $f$ is an isometry. Then a connected component $V$ of $\text{Fix}(f)$ is a closed, totally geodesic submanifold without boundary (Fixed Points Set of an Isometry), and contains w.l.o.g. by assumption an open subset of $U$. Therefore $V$ is open (as a submanifold without boundary of $\dim V=n=\dim U$) and closed. Since $U$ is connected, $V=U$. So $U=V\subseteq \text{Fix}(f)=U$ and $f=\text{id}$.
If $X$ is a connected topological manifold and $f$ only continuous this still holds: $\mathbb{Z}_n \to \text{Homeo}(X), 1 \mapsto f$ is an effective action, if $n$ is minimal with $f^n = \text{id}$. By one of Newman's theorems on transformation groups $N = \{x\in X: (\mathbb{Z}_n)_x=1\}$ is open and dense in $X$. Therefore it exists some $x \in \text{Fix}(f) \cap N \neq \emptyset$ and therefore $1=(\mathbb{Z}_n)_x = \mathbb{Z}_n$, so $n=1$ and $f=\text{id}$.