When solving a set-theoretic problem, I needed to use the following claim:
Claim: For every infinite set $A$, there exists a bijective function $f:A\rightarrow A$ with no fixed points (that is, $f(a)\neq a$ for all $a$).
(The analogous claim for finite sets is quite easy to prove.)
I proved it using the observation that $|A|=|A\times \{0,1\}|$, and constructing a bijective function $g:A\times \{0,1\} \rightarrow A\times \{0,1\}$ with no fixed points: $$g(a,0)=(a,1) , g(a,1)=(a,0).$$
However, some variant of choice is needed in order to show that $|A|=|A\times \{0,1\}|$.
Is there a way to prove my claim without using choice? My intuition tells me it's impossible. If so, can one show that the claim's negation is consistent with ZF?
We say that $A$ is a strongly amorphous set if it is infinite and whenever we partition $A$ there is at most one infinite part, and all but finitely many parts are singletons.
If $f\colon A\to A$ is a permutation, then we can decompose $A$ to the different orbits. But we can show that no orbit is infinite, since that would imply a copy of the natural numbers can be found in $A$ and it is not strongly amorphous. Therefore there are infinitely many orbits, so most of them are singletons. In other words, only finitely many points even move at all.
Strongly amorphous sets have analogous for larger cardinals $\kappa$ which are consistently existing while $\sf DC_{<\kappa}$ holds. And similar arguments can be made in that case.