Any model of $ZF$+stationary proper class of $I3$ ordinals (whose consistency strength is at most $ZF$+$I2$) where $W_α$ is listing of $V_{κ}$ where $κ$ is $I3$ and $j_α$ the witness of $I3(V_κ)$ will be a model of your theory.

Injectivity and Cumulation are trivial, $ZF$+Elementarity is part of our assumption.

For Reflection, let $φ$ be any formula, by Levy's reflection principle there is a proper class club $C_φ$ that reflect $φ$ (note that Levy's reflection works even with a modified language).

Because there is a stationary set of $I3$ cardinals, there exists a $I3$ cardinal in $C_φ$, then this ordinal will witness Reflection of $φ$.

This section is a follow-up based on the comments.

Let $\cal L$ be the language of our theory (or any other language extending $\{∈\}$), and let $ZF_{
\cal L}$ be $ZF$ with schema over the language $\cal L$ then *any model of $ZF$ is (under a suitable interpretations) a model of $ZF_{
\cal L}$*.

The reason for this is that we do not have any axioms to restrict
any of our new symbols, let interpret each relation symbol in $\cal L\setminus\{∈\}$ as a tautology, each function symbol as the identity, and each constant as the emptyset.

Because all 3 of those are definable, any formula containing any symbols in $\cal L$ will be equivalent to a formula in $\{∈\}$.

So $\forall \alpha \,( W_\alpha \models \sf ZF_{\cal L})$ doesn't imply $W_\alpha$ is a model of ZF+Reinhardt, because while it does satisfy schema over the symbol $j$ and $W$, it does not see any nee from the universe to itself (The symbol of $j$ in $W_α$ **has no relation** to the symbol $j$ in $V$)

Now what about the reflection principle? In our language we have $2$ extra symbols, $j,W$.

We can restrict ourselves to $α$ such that $W_α$ reflects enough to prevent the triviality we had in the second paragraph of the follow-up, but it doesn't matter.

To show that it doesn't matter, look at the sentence $∀x(x∈Ord\implies ∃y∈Ord(\text{such that $W_x=V_y$ and $j_x$ is nee from $W_x$ to itself}))$, assume it is reflected into $W_α$, why is $W_α$ not Reinhardt? The reason is that $W_α$ does not see all of the ordinals, so while we may have $α∈W_α$ (well, for this specific sentence if it is reflected to $W_\alpha$ then $V_\alpha=W_\alpha$ so $\alpha\notin W_\alpha$, but this is irrelevant to my point), the value of $W_α^{W_α}$ is a **proper** subset of $W_α$, so $j,W$ inside of $W_α$ cannot say anything about the real $W_α,j_α$, i.e. we can not see any nee on $W_α$ inside of $W_α$.

So **how can we get Reinhardt**? To get Reinhardt we **must** adjust the nee $j_α$ to $W_α$, define for a set $X,Y$ the model $(X,∈,Y^{(k)})$ be the model $(X,∈,Y)$ but the symbol for $Y$ is $k$ and add the following axiom (remember that as long as we don't add interpretations/axioms on new symbols, they don't change the theory, so we can require just $ZF_{\{∈,k\}}$ instead of $ZF_{\{∈,k,j,W\}}$):

For every ordinal $α$, $(W_{α},∈,j_{α}^{(k)})⊨ZF_{\{∈,k\}}$

This is exactly your theory with the extra axiom that "$W_α$ can see the nee on itself".

This theory is very strong, I *want to say* that this theory is equivalent to "ZF + there is stationary proper class of $α$ such that $V_α$ is a model of Reinhardt", but I am not 100% sure about the $⇒$ direction (the proof of $⇐$ direction is essentially the same as my original answer but replacing $I3$ with Reinhardt).

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