TITLE: Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?)
QUESTION [21 upvotes]: Let us say that a set B admits a rigid binary relation, if there is a binary relation R such that the structure (B,R) has no nontrivial automorphisms. 
Under the Axiom of Choice, every set is well-orderable, and since well-orders are rigid, it follows under AC that every set does have a rigid binary relation. 
My questions are: does the converse hold? Does one need AC to produce such rigid structures? Is this a weak choice principle? Or can one simply prove it in ZF?
(This question spins off of Question A rigid type of structure that can be put on every set?.)

REPLY [4 votes]: A variant of Justin Palumbo's answer, blackboxing the forcing construction: 
An infinite set X is called amorphous if all its subsets are finite or co-finite.  X is called strongly amorphous (or superamorphous) if every relation on X (that is, every subset of $X^n$, for any $n$)
is definable with finitely many parameters in the language of equality. (In other words, restricting to $n=2$ for simplicity: Every $R \subseteq X^2$ must be in the Boolean algebra generated by the sets $X_a:=\{(a,x): x\in X\}$, their converses and the diagonal.  Unless I have forgotten a few more generators.) 
(In particular, amorphous sets are Dedekind finite.) 
Clearly, strongly amorphous sets have no rigid relation.  Their consistency with ZF can be shown (as pointed out above) by first constructing a ZFA model with an amorphous set of atoms, then applying the Jech-Sochor theorem.  
(Strongly) amorphous sets X (and related structures, such as the powerset of X) can often be used as counterexamples showing that some choice is needed.