Consider the following statement in $\sf ZF$:



> 
> (I) Whenever $X$ is a set with more than $1$ element, there is an injective map $\iota: X\to X$ such that $\iota(x) \neq x$ for all $x\in X$.
> 
> 
> 


The [Axiom of Choice (AC)](https://en.wikipedia.org/wiki/Axiom_of_choice) implies (I) -- but does (I) imply (AC)?



Injections without fixed-points and the Axiom of Choice
It is shown in 
Tachtsis, E.  

On the existence of permutations of infinite sets without fixed points in set theory without choice.  

Acta Math. Hungar. 157 (2019), no. 2, 281-300.
that ZF+(every infinite set supports a permutation with no fixed points)
does not imply AC. It is easy to see in ZF that a finite set supports a cyclic permutation, so that should be sufficient for your question.






