Every set with more than point admits a permutation with no fixed point and the Axiom of Choice

Question

Assuming axiom of choice , for any set $S$ with more than one point , there exist a bijection $f:S \to S$ such that $f(s) \ne s , \forall s \in S$ . Is the converse true , i.e. Does the statement " For every set $S$ with more than one point , there exist a bijection $f:S \to S$ such that $f(s) \ne s , \forall s \in S$ " implies Axiom of Choice ? Or at least can we prove that any set with more than point has a permutation with no fixed point , without axiom of choice ? Please help. Thanks in advance

Answer 1

This statement allready follows from the addition theorem, which states that $A \times \{0,1\}$ is equipotent with $A$ for every infinite $A$ (as on $A \times \{0,1\}$, $(a,x) \mapsto (a, 1+x)$ is fixed-point-free). The addition theorem is known to be strictly weaker than the axiom of choice (a result by Sageev).