Suppose we have sequences $\zeta_i^n$ with supremums $\lambda_n$, and elementary embeddings $e_{i,j}^n: V_{\zeta_i^n}\prec V_{\zeta_i^n}$ with critical point $\kappa_i$. The embeddings $e_{i,i+1}^n$ generate embeddings $j_n: V_{\lambda_n}\prec V_{\lambda_n}$: If $\zeta_j^n\lt rank(x)\lt\zeta_i^n$ for $j\lt i$, $j_n(x)=e_{i,i+1}^n(x)$. Assume that $j_n$ extends in the usual manner to an I1 embedding $j_n^+: V_{\lambda_n+1}\prec V_{\lambda_n+1}$.
Now define the embedding $e^{(i)}=e_{i,i+1}^i$. Let $\lambda=\text{lim}_{n\lt\omega}\zeta_n^n$. If $\zeta_n^n\lt\zeta_{n+1}^{n+1}$ for every $n$, does it follow that the embeddings $e^{(i)}$ induce an I1 embedding? If the respective $j^n$ extend to an $I0$ embedding, does it follow that the $e^{(i)}$ induce an $I0$ embedding?
A few minor modifications to the conditions are allowed.
Motivation: I was doing research on large cardinals beyond choice, and I was able to construct an argument that from some $j: V\prec V$, you can get a transitive model of $ZFC$ with a proper class of $\omega$-huge cardinals, assuming this is true. If the embeddings do in fact extend to $I0$ embeddings, I think we can get a model of $ZFC$ in which there is a proper class of $I0$ cardinals. In fact, I think that with a few minor modifications to the argument, you can get away with just some $j: V_{\lambda+2}\prec V_{\lambda+2}$.
Edit: Looking at the construction, I think the embeddings $e^{(i)}$ do induce an $I3$ embedding. Let $e$ be the induced embedding. Note that in the proof that if $j: M_0\prec_1 M_1$ if and only if $j: M_0\prec M_1$ (In The Higher Infinite), he uses exclusively the fact $j(V_\alpha^{M_0})=V_{j(\alpha)}^{M_1}$. But this is obvious for $j=e$.