@AsafKaragila it is not exactly "a lot of I3 cardinals", as "Elementarity" requires all of the elementary embeddings to extends each other

@ZuhairAl-Johar I am using the fact that "being an ordinal" is absolute between transitive models, you don't need them to be models of ZF ("$x$ is an ordinal" is $\Delta_0$), but yes, I did use that fact. With a bit more effort you can show that even if we don't assume $W_0$ is transitive your strong reflection principle implies "being is ordinal" is absolute between $V$ and $W_0$

Zuhair, your edit to your question has change it immensely, I'll add an answer for the edited question but for the future if you have "small edits" like those, ask in the comment of the existing answer and see if the OP of the answer will have an answer or it deserves a different question, you should avoid editing questions as much as possible

@ZuhairAl-Johar I edited my answer, I hope it helps

@ZuhairAl-Johar I will edit soon today (or tomorrow morning) to write a fully detailed explanation about all of the discussion in the comments inside of the post, and I hope it will clear things up

@ZuhairAl-Johar $W_{α+1}$ can see that the $κ$ such that $W_α=V_κ$ is $I3$, but for it to be Reinhardt you need to have that $W_α$ internally see the replacement of $j_{α}$

@ZuhairAl-Johar If you want $(W_α,∈,j_α)$ to be a model of $ZF+Reinhardt$ you need $j_{α}^{(W_α,∈,j_α)}$ to be elementary embedding, there is no reason to believe that $j_α^{(W_α,∈,j_α)}$ have anything related to $j_α^V$. To maybe better understand this, try to think about $j_{2^{2^{2^{|α|}}}}^{(W_α,∈,j_α)}$, it can't be $j_{2^{2^{2^{|α|}}}}$ because $j_{2^{2^{2^{|α|}}}}$ is a function between $W_{2^{2^{2^{|α|}}}}$ to itself, but $W_α$ doesn't see anything close to $W_{2^{2^{2^{|α|}}}}$

@HanulJeon no worries, with the additional context your comment is on point

@HanulJeon "separation and replacement for formulas with $j_α$" still doesn't mean $W_α$ will interpret $j_α$ as elementary embedding, I don't understand how separation and replacement for $j_α$ will do anything without any extra (non-ZF related) axioms

@HanulJeon Requiring $\forall \alpha \,( W_\alpha \models \sf ZF)$ with the new language does not makes each $W_α$ satisfy $ZF+j_α\text{ is elementary embedding}$, $W_α$ may interpret $j_α$ different from how $V$ interpret it

@ZuhairAl-Johar as FarmerS said, just requiring it being a model of $ZF$ doesn't add any power, you need to require something on how it interpret $j_α$ to change anything

@HanulJeon it does not matter how to formulate it, to get Reinhardt you need "ZF+(schemas including j)+**j is elementary embedding**", just adding symbols without any restriction on the symbols doesn't change anything (let $L=FOST + F + R + C$ where $F$ are set of new function symbols, $R$-set of relations symbols, $C$ set of new constants, interpret every new symbol as $∅$, and any model of ZF will be a model of ZF with the new language with schemas allowing the new symbols)

@ZuhairAl-Johar No, it is much weaker. The problem is that having a lot of elementary embedding doesn't give us a way to "stich them together" to get Reinhardt. And even if you had a way to stich them together there is no reason to believe that the "resulting stich" won't be the identity. (For example, it is consistent [relative to blah blah] that there is proper class of $I3$ ordinals, and for every ordinal $α$, there is some $β$ such that the elementary embedding for the $I3$ ordinals above $β$ are the identity on $V_α$, so the embeddings are "eventually the identity")