Does every set with at least two or more elements admit a fixed-point free self-bijection? Assume $X$ is a set with at least two elements. Is there a bijection from $X$ to $X$ such that for every $x\in X$ , $f(x) \ne x$ ?
 A: Yes, at least if you allow some choice.
Let $\mathcal T$ be the collection of bijections $f\colon U\to V$ where $U,V$ are disjoint subsets of $X$. Then $\mathcal T$ is not empty (it contains $\operatorname{id}_\emptyset$) and is inductively ordered in the obvious way (i.e., by letting $(f\colon U\to V)\ge (f'\colon U'\to V')$ iff $U'\subseteq U$ and $V'\subseteq V$ and $f|_{U'}=f'$). Hence by Zorn's lemma, there is a maximal element $m\colon A\to B$ of $\mathcal T$.


*

*If $A\cup B=X$ we are done as we can let $$\phi(x)=\begin{cases}m(x)&x\in A\\m^{-1}(x)&x\in B\end{cases}$$

*If $X\setminus(A\cup B)$ has at least two elements $a,b$, we can extend $m$ to $\tilde m\colon (A\cup \{a\})\to (B\colon \{b\})$ by mapping $a$ to $b$. As this would contradict the maximality of $m$, this case does not occur

*If $X\setminus(A\cup B)=\{c\}$ we can proceed as follows: As $X$ has at least two elements, $A$ and $B$ cannot be empty.  Pick $a\in A$ and define
$$ \phi(x)=\begin{cases}m(x)&x\in A\setminus\{a\}\\
m^{-1}(x)&x\in B\\
m(a)&x=c\\
c&x=a\end{cases} $$ and that is our bijection.

A: Outline: Let $X$ be an infinite set, and let $Y$ be a set disjoint from $X$ which has the same cardinality as $X$. Then it is a standard result of ZFC that $X\cup Y$ has the same cardinality as $X$. Now it is easy to find a bijection from $X\cup Y$ to $X\cup Y$ that has no fixed point. Just map any element of $X$ to its "twin" in $Y$, and vice-versa. 
A: Here’s a counterexample in the absence of the axiom of choice.
Suppose that $X$ is strictly amorphous, and suppose that $f:X\to X$ is a bijection. Since $X$ is Dedekind finite, there is no injection from $\Bbb N$ to $X$, and therefore for each $x\in X$ there is a least integer $n_x\in Z^+$ such that $f^{n_x}(x)=x$. It follows that each orbit under $f$ is finite. The orbits under $f$ partition $X$, so infinitely many of them are of size $1$, and $f$ has infinitely many fixed points.
