I'm curious if the following property always holds. Let $f:X\rightarrow X$ be a bijection from an infinite set $X$ into itself, such that for all $x\in X$ it holds that $f(x)\neq x$. Then there exists some subset $A\subseteq X$ such that $$X=A\sqcup f(A). $$ Where $\sqcup$ means that the union is disjoint.
A simple example of this property is provided by the map $f:\mathbb{Z}\rightarrow\mathbb{Z}:n\mapsto n+1$, where the corresponding subset of $X=\mathbb{Z}$ is $A=2\mathbb{Z}$.
Is this property always true? Can someone provide a counter-example?
No. Odd cycles are your enemy.
Take $X=S^1$ and $f(z)=e^{\frac 23\pi i}z$. Then $f(f(f(x)))=x$ for all $x$ and you have problems assigning $x,f(x),f(f(x))$ to $A$ without conflicts.
Smaller counterexample: $X=\Bbb Z$ and $$ f(x)=\begin{cases}x+1&x\in\{-1,0\}\\-x&\text{otherwise}\end{cases}$$