I am not sure my proof is correct/complete for the following:
For a function $f : A \to A$ denote by $f^n$
the composition $f \circ \cdots \circ f$ ($n$ times). Show that if for each
$x\in A$ there is an $n(x) \in \mathbb N$ such that $f^n(x) = x$, then $f$ is bijective.
Proof: We see that $f \circ \cdots \circ f(x) = x$ is a linear and thus a bijective function.
Notes:
Given 2 functions $f$ and $g$, if $g \circ f$ is injective, then $f$ is injective.
Also, if $g \circ f$ is surjective, then $g$ is surjective.
Therefore,
$f^n = f^{n-1} \circ f = x$ is bijective therefore injective and thus $f$ is injective for every given $x$ in $A$
$f^n = f \circ f^{n-1} = x$ is bijective and thus surjective and by the notes above, $f$ is surjective for every given $x$ in $A$
We conclude $f$ is bijective.
Best Answer
Proof. The function $f$ is surjective because $f(...(f(x))...)=x$ for every $x$. Suppose that $f$ is not injective, $f(a)=f(b)$ for some $a\ne b$. We have that $f^n(a)=a, f^m(b)=b$.Then $f^{mn}(a)=a\ne b=f^{mn}(b)$. But then $a=f^{mn}(a)=f^{mn-1}(f(a))=f^{mn-1}(f(b))=f^{mn}(b)=b$, a contradiction.
Hence $f$ is bijective.