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

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

marked as duplicate by Asaf Karagila♦ axiom-of-choice StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 13 '15 at 15:25

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).