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 appeared on the arXiv
proposing a resolution of that longstanding open question. But the paper has yet to be fully vetted by the community.)
