EDIT: it appears that my original question has some confusion between auto-morphisms and elementary embeddings as it is obvious from the answer below, therefore I'll clarify here what I exactly want.


Can we have the following?


A transitive model $M$ of ZF-Regularity such that we have an external injective function $j$ from $P(M)$ to $P(M)$ such that $j$ is an ismomorphism on $\in$ between $P(M)$ and $range(j)$. Now we demand that $M$ is a subset of $range(j)$, and that every definable subset of $M$ [from parameters in $M$] would be sent by $j$ to an element of $M$. 


Where "$\kappa $ is a definable subset of $M$ from parameters $w\_1,..,w\_n \in M$", is defined as: $$ \forall y (y \in \kappa \leftrightarrow y \in M \wedge \phi(y,w\_1,..,w\_n))$$ for some formula $\phi(y,w\_1,..,w\_n)$ that is written in the language of set theory [with the only exception of allowing $j $ to be appear on parameters, so $j(w\_i), j^{-1}(w\_i)$ are allowed].


We know that $range(j) \neq M$, because the parity of $M$ is different from $P(M)$.


The question is if there is a clear inconsistency with that?




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Is the following known to be consistent with some extension of $\text{ZF}$?


There is a transitive model $M$ of $\text{ZF}$ such that there is an external non-trivial elementary embedding $j$ from $P(M)$ to $P(M)$ (where $P$ is the known Power operator) such that $M \subset range(j)$ and such that every definable subset $\kappa$ of $M$ (parameter free or from parameters in $M$) in the language of set theory (i.e. doesn't use the symbol $j$) is sent by $j$ to an element of $M$[i.e. $j(\kappa) \in M$] and such that the symbol $j$ can be used freely in the instances of Replacement and Separation. 


To clarify what is meant by "$\kappa $ is definable subset of $M$ from parameters $w\_1,..,w\_n \in M$", is to mean that: $$ \forall y (y \in \kappa \leftrightarrow y \in M \wedge \phi(y,w\_1,..,w\_n))$$ for some formula $\phi(y,w\_1,..,w\_n)$ that is written in the language of set theory.


In particular: is the above consistent with $\text{ZF + Reinhardt cardinal}$?



Can there be such an elementary embedding?
EDIT: Based on the comments, I think it's worth clarifying a bit of the nature of the Boffa-Jensen construction of a model of NFU (which appears to be part of the motivation for this question).


In this construction, we do not have an elementary embedding; rather, we begin with a model $M$ of ZFC and an *automorphism* - unfortunately for our context denoted "$j$" in [Holmes' article](https://math.boisestate.edu/~holmes/preserves3.pdf), but which I will call "$a$" - which moves a rank, that is, which is not the identity on the ordinals. I think the conflation of automorphism and elementary embedding is generating some confusion here.


*Incidentally, a fun exercise is to show that any nontrivial automorphism of a model of ZFC must move an ordinal, while this is not true of ZF! I believe this was first observed by Harvey Friedman. A cute result of Cohen is [the construction of a (necessarily ill-founded) model of ZF with a nontrivial automorphism of finite order](https://www.ams.org/books/pspum/025/pspum025-endmatter.pdf), and it was in Cohen's paper that I first saw this fact quoted.*


Now we come to a key point: the automorphism $a$ is completely *external*, and the mere existence of a nontrivial automorphism immediately implies that $M$ is illfounded. Put better: any well-founded model of ZFC, or ZF, or even much weaker, has no nontrivial automorphisms. *(Think about the least rank of an element moved by a hypothetical nontrivial automorphism …)* Indeed, when we speak of automorphisms of models of set theory these are *always* understood as being completely external to the models in question; note that this is in contrast with the situation for elementary embeddings, where there are various flavors of definability. Similarly, the connection with ill-foundedness breaks down for elementary embeddings: assuming $0^\sharp$ exists, there is a nontrivial elementary embedding of $L$ into itself, but $L$ is of course well-founded. Automorphisms and elementary embeddings are really very different objects.


Anyways, in the Boffa-Jensen construction we take an $M$ and $a$ as above, and fix some $M$-ordinal $\alpha$ with $a(\alpha)\not=\alpha$. WLOG we have $a(\alpha)<\alpha$ (otherwise replace $a$ with $a^{-1}$. The structure $\mathcal{B}$ associated to the data $(M, a, \alpha)$ has domain $(V\_\alpha)^M$, and elementhood relation $\varepsilon$ given by $$x\varepsilon y\iff a(x)\in^My\wedge y\in^MV\_{a(\alpha)+1}.$$ Everything in $V\_\alpha^M\setminus V\_{a(\alpha)+1}^M$ winds up being an urelement in the sense of $\mathcal{B}$: if $y\in V\_\alpha^M\setminus V\_{a(\alpha)+1}^M$, then we never have $x\varepsilon y$ for any $x\in V\_\alpha^M$.


*(Why is $V\_\alpha^M\setminus V\_{a(\alpha)+1}^M$ nonempty? Well, one way to guarantee this would be to make sure that $M$ is an $\omega$-model, that is, the "finite ordinals" in the sense of $M$ are in fact all finite; however, even this assumption is unnecessary, since $a(\alpha)$ and $\alpha$ must have the same "$M$-parity" and no ordinal has the same parity as its successor.)*


However, we're seeing something very specific to *automorphisms* here. An elementary embedding $M\preccurlyeq M$ simply wouldn't help us do the same thing: it is crucial that $a$ moves ranks *downwards*. Having an "upward-moving" automorphism is still fine, since we can invert it, but elementary embeddings aren't surjective in general so no similar trick will work there. Indeed, I know of no way to get a model of NFU (or similar) from, for example, $0^\sharp$ in a natural way (remember that $0^\sharp$ is equivalent in a sense to a nontrivial elementary embedding from $L$ into itself; this isn't a contradiction with Kunen, since "$0^\sharp$ exists" implies $V\not=L$).




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It's not clear to me exactly what structure you're looking at, but I suspect it's (equivalent to) the $\{\in\}$-structure $\mathfrak{S}=M\sqcup P(M)$. *(If not, what do you mean?)* Under this interpretation the answer to your question is **no**. Consider $M$ as an element of $\mathfrak{S}$. $M$ is certainly a definable subclass of $M$, so we must have $j$ send $M$ to some $a\in M$; in particular, $j(M)\not=M$ (where again we're thinking of $M$ here as an *element of* our "power structure" $\mathfrak{S}$, not as a *subset of* it).


But $M$ is a definable element of $\mathfrak{S}$: $M$ is the unique $z\in\mathfrak{S}$ such that for all $a\in\mathfrak{S}$ we have $\mathfrak{A}\models\exists x(a\in x)\iff a\in z$. So $M$ can't be moved by an elementary embedding. 




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As far as I can tell, this argument (or a version of it) works for any reasonable envisioning of $P(M)$ as a structure. For example, another possible interpretation of your question eschews "$\in$" but has the same result. We look at a structure $\mathfrak{S}$ whose elements correspond to subsets of $M$ (if you want, the domain of $\mathfrak{S}$ is exactly $P(M)$). The language of $\mathfrak{S}$ is just $\{\subseteq\}$, interpreted in the obvious way.


The domain of $\mathfrak{S}$ has a special subset $Ground$: namely, the set of all $a$ with $\{x\in M: x\in a\}=\{x\in M: M\models x\in b\}$ for some $b\in M$. These are the "internal sets" of $M$. 


The first two requirements you put on $j$ are elementarity and $j(a)\in D$ whenever $a$ is a definable subset of $M$. The latter means in particular that we must have $j(M)\in D$ (note that $M\in \mathfrak{S}$). But then $j$ certainly can't be elementary, since $M$ is the unique $a\in\mathfrak{S}$ satisfying $\forall x(x\subseteq M)$ (and elementary maps can't move definable elements.


