TITLE: How much choice follows from the existence of derangement?
QUESTION [9 upvotes]: We can show that every set has a derangement (that is, a bijection $f$ from itself to itself such that $f(x)\neq x$ for all $x$) if we assume the axiom of choice. In fact, the full axiom of choice is not necessary: if $|A+A|=|A|$ for all $A$, then we can find a derangement.
Moreover, some kind of axiom of choice is necessary to prove it. (According to the linked answer, $\mathsf{AC}_\kappa$ and $\mathsf{DC}_\kappa$ does not suffice to provide a derangement for any set.) 
My question is: what kind of choice axioms follow from the existence of derangement for all sets? What I can show is the following simple fact:

If every set has a derangement, then either every set has a countably infinite subset or there is a set $A$ such that every derangement $f$ over $A$ satisfies $\forall x\in A\exists n<\omega: f^n(x)=x.$
(The proof is easy: if every set has an element which satsify $f^n(x)\neq x$ for all $n<\omega$, then it witnesses a countably infinite subset.) 

This fact has a choice-related aspect (i.e. every infinite set has a countably infinite subset) but also has a counterweight for the choice (i.e. every derangement does not have an element of an infinite order w.r.t. $f$.)
I wonder there is a known choice principle follows from the existence of derangement. I would appreciate your help!

REPLY [2 votes]: One can generalize the fact you stated in the midde using other finiteness definitions:

If $A$ is infinite set with $f$ derangement then:

If the cycles are not a partition of $A$, then $A$ is Dedekind infinite.

If the cycles are a partition of $A$ and the size of the sets in the partition is unbounded then $\mathcal{P}(A)$ is Dedekind infinite.($Δ_4$-infinite by Truss notation)

If they are a partition and the size of the elements of the partition are bounded then $A$ is not strictly amorphous.




You showed the first case, now if $P$ is the partition generated from the cycles, and $\{n∈ω\mid ∃x∈P(|x|=n)\}$ is indeed infinite, one can just take the function $g:ω→{\cal P}(A)$ to be $g(n)=\bigcup\{a⊆A\mid a\in P\mbox{ and }|a|=n\}$, then by removing $g^{-1}(0)$ from the domain we get that $ω<|\mathcal P(A)|$.
The definition of strictly amorphous is an amorphous set with gauge equal to $1$, in other words, for every partition of $A$ we have finitely many non-singeltons, which implies every bijection on $A$ will have a fixed point.

Note that it is possible for amorphuous set to have derangement, for example if it's gauge is $2$, then $A$ is weakly even(we can partition it to subsets of size 2), and just map each element in $A$ to the unique element that is with it in the partition.
And because $\kappa$ has derangement $⇒$ $\kappa+1$ has derangement(if there exists a cycle you can just modify a single cycle to include the new element, if there are no cycles just insert it in one of the $\Bbb Z$-chains), we must have that weakly odd sets also have a derangement.
I think that it is consistent that there exists amorphous set and that all amorphous sets are either weakly even or weakly odd(please correct me if I am wrong), in that case it is consistent that all amorphuous sets has derangement.
It is also consistent that there exists an infinite Dedekind finite set and that the class of Dedekind finite cardinals is linearly order, a consequences of that is that the class of Dedekind finite cardinals is a model of arithmetic(if someone has reference to this I would love to get it), and so every Dedekind finite cardinal is either odd or even, so it is either weakly odd or weakly even, hence it is consistent that all Dedekind finite cardinals have a derangement.