When atomhood is definable, the answer is clearly yes, so essentially the question is interesting primarily when we are working in ZF with extensionality weakened to apply only to inhabited sets. In this case, we may as well include the empty set with the atoms. So, if we have an elementary embedding from $V\_0$ to $V\_0$, we can turn this to an elementary embedding from $V$ to $V$. Obviously, an elementary embedding from $V\_0$ to $V\_0$ is just a self-injection. (I'm being a bit careless, since $V\_0$ might not be a set.)


My question is then, do all the elementary embeddings arise this way?



Does every elementary embedding $j:V \to V$ in ZFA arise from a self-injection on the class of atoms?
In ZFCA, the answer is yes, every elementary embedding $j:V\to V$
is the unique extension to $V$ of an injection on the atoms. If the
class of atoms is a set, then it must be a permutation of the
atoms.


On the one hand, every injection $\pi:A\to A$ on the class $A$ of
atoms extends naturally to a map defined on all of $V$ by defining
recursively $j\_\pi(u)=\{j\_\pi(x)\mid x\in u\}$ for any set $u$. This
map will fix all pure sets and $j\_\pi$ will be an isomorphism of
$V=V(A)$ to the universe $V(\pi"A)$ built using atoms in the range
of the original map. One can show that that structure is an
elementary substructure of the universe when the class of atoms is
a proper class. If $A$ is a set, then any permutation of $A$
extends to an automorphism of the universe $V$ by similar means.


Conversely, suppose that $j:V\to V$ is an elementary embedding. Let
$\pi=j\upharpoonright A$ be the action of $j$ on the class of
atoms. I claim that $j=j\_\pi$. To see this, notice first that by
restricting $j$ to the pure sets, those with no atoms in their
transitive closure, we must get the identity embedding, because of
the original ZFC version of the Kunen inconsistency. In particular,
$j$ fixes every ordinal. It follows that $j(w)=j"w$ for any set of
atoms, since we can well-order the set $w$ and then observe that
$j$ must carry the $\alpha^{th}$ element to the $\alpha^{th}$
element and the length of the sequence does not get longer.


For any set of atoms $w$, we define the rank hierarchy by


* $V\_0(w)=w$
* $V\_{\alpha+1}(w)=P(V\_\alpha(w))$
* $V\_\lambda()=\bigcup\_{\alpha<\lambda}V\_\alpha(w)$, at limits


Rank can also be defined by $\in$-recursion, and ZFCA proves that
every set $u$ is in some $V\_\alpha(w)$, where $w$ is the set of
atoms in the transitive closure of $u$.


What I claim is that $j\upharpoonright V(w)$ is an isomorphism of
$V(w)$ with $V(j''w)$. One can simply argue by induction on rank
that $j$ carries $V\_\alpha(w)$ to $V\_\alpha(j"w)$, and
furthermore, that it does so in the same way that $j\_\pi$ does. If
this is true at $\alpha$, then it is true at $\alpha+1$, using the
fact that $j(V\_{\alpha+1}(w))=V\_{\alpha+1}(j(w))$, which follows
from the fact that $j(\alpha)=\alpha$.


Thus, we conclude that $j=j\_\pi$, and so yes, every elementary
embedding from $V$ arises as the unique extension to $V$ of a map
on the atoms.


In ZFA, without the axiom of choice, the question is open, since it contains the ZF Reinhardt cardinal as a special case. In the ZFCA argument above, we used the axiom of choice in order to know
that $j$ fixes every ordinal, and also to know that $j(w)=j"w$ for any set $w$ of atoms. Without the axiom of choice, the
question reduces to the Reinhardt cardinal question when there are no atoms. 


(One might mention that just a few days ago, [a preprint by Rupert
McCallum](https://arxiv.org/abs/1712.09678) appeared on the arXiv
proposing a resolution of that longstanding open question. But the paper has yet to be fully vetted by the community.)


