Although any permutation of atoms induces an automorphism of the whole universe, atoms seem to be indiscernible only within the permutation models. Can a permutation model be extended to a rigid structure? We can suppose that AC holds.
1 Answer
You have answered your own question.
In ZFA, every definable permutation $\pi:A\to A$ of the class of atoms extends uniquely to an automorphism of the entire settheoretic universe, simply by defining it on sets $y$ as follows: $$\pi(y)=\{\pi(x)\mid x\in y\}.$$ It is easy to see that $x\in y\iff \pi(x)\in \pi(y)$, and so this is an isomorphism. (One needs to know that $\pi$ is definable in order to know that the extension is definable, which is needed in order to know that $\pi(y)$ is a set, etc.)
If there are at least two atoms, then we can define $\pi$ so as to swap them, fixing all other atoms, and this map will extend to a definable automorphism of $V$. So in ZFA, if there are at least two atoms, then the universe is not rigid.

1$\begingroup$ This doesn't seem to answer the question. The OP was asking about extensions that can impose rigidity, he suggested AC, but I think this is not enough, we need GC to make a theory with atoms rigid. $\endgroup$ Feb 8 at 22:07

4$\begingroup$ Yes, I had answered the question asked in the title "Is the universe of ZFA rigid". The OP seems to have asked a different question in the body of the post. But actually, I am unsure what that question in the body is asking. The claim that atoms are indiscernible only within the permutation models is incorrect. Also, under AC every setsized structure can be trivially extended to a rigid structure, but it is unclear whether that is the kind of answer that is sought. $\endgroup$ Feb 8 at 22:22

2$\begingroup$ Thank you, Joel and Zuhair for the answers. Perhaps I was not clear in my question. In other words: under AC in ZFA, does Joel and Palumbo's result that every set has a rigid relation also holds, that is, even for atoms? I guess that even a permutation model, which is a structure in ZFA, can be extended to a rigid structure. Is this correct? $\endgroup$ Feb 8 at 23:44

$\begingroup$ Sorry to insist, but I am in doubt about something you don't find in the books and seems to require the interpretation of a more experienced set theorist. In ZFA we have the pairing axiom, so we can form the unitary set {a} even if a is an atom. Why this fact would not tell us that the atoms can always be discerned from one another? Fraenkel took a collection of "distinct" atoms for his "cells". Then, under extensions of permutations, the atoms "are made" indiscernible within the model. But I think that they are not for someone living outside the Model. Any mistake in this reading? $\endgroup$ Feb 9 at 11:30

$\begingroup$ In ZFA atoms are indiscernible in the sense that every assertion $\varphi$ expressible in the language of set theory that holds of one atom $\varphi(a)$ holds of all atoms $\varphi(b)$. You cannot tell them apart in the language of set theory. This is because there is an automorphism swapping $a$ and $b$, and automorphisms preserve truth. $\endgroup$ Feb 9 at 12:58