$\DeclareMathOperator{\crit}{\operatorname{crit}}$A rank-into-rank embedding is a non-trivial elementary embedding from a rank initial segment of $V$ into itself: $j:V_\delta\prec V_\delta$. Define the critical sequence of such an embedding by setting $\kappa_0=\crit(j)$ (the first ordinal moved by $j$) and $\kappa_{n+1}=j(\kappa_n)$. Let $\lambda=\crit^\omega(j)=\sup_{n<\omega} \langle \kappa_n\rangle$. It is straightforward to see that $\lambda$ is a strong limit cardinal of countable cofinality.

By a theorem of Kunen, if such an embedding can exist, then $\delta$ must be the ordinal $\lambda$ or $\lambda+1$.

It is not hard to see that $\crit(j)$ must be measurable. In fact, for any $n$, $\crit(j)$ is also $n$-huge as witnessed by the ultrafilter $$U=\{X\subseteq\mathcal{P}(\kappa_n): j"\kappa_n\in j(X)\}.$$ Further, if we let $j^n$ denote $j$ composed with itself $n$ times, then $$V_\lambda\models ``\lambda\text{ is supercompact"}.$$ To see this, suppose $\crit(j)\leq \theta <\kappa_n$, then $$U=\{X\subseteq\mathcal{P}_{\crit(j)}(\theta): j^n"\theta\in j^n(X)\}$$ winesses the $\theta$-compactness of $\crit(j)$ (in $V_\lambda$).

For the last claim, it is enough that $\crit(j)$ is $<\lambda$-supercompact, i.e. not *fully* supercompact in $V$. In this case, however, $\crit(j)$ *could* be fully supercompact.

But extendible cardinals are not characterized by the presence of ultrafilters and this motivates my question here.

**Question:** Can the critical point of a rank-into-rank embedding be extendible?

It may not make sense (I think) to ask for full extendibility of $\crit(j)$: Suppose otherwise that $\crit(j)$ is fully extendible. Let $k$ witness the $\theta$-extendibility of $\crit(j)$ for some $\theta>\crit^\omega(j)$. Then we have $$V_{\crit(j)}\prec V_{\crit^\omega(j)}\prec V_\theta.$$

This looks suspiciously like Woodin's Enormous Cardinal (though his notion is defined in the context of just ZF). See http://logic.harvard.edu/EFI_Woodin_talk.pdf, slide 20. Thus I'm not sure that $\crit(j)$ can be fully extendible.

**Question:** Assume $j$ is a rank-into-rank embedding and let $\lambda=\crit^\omega(j)$. Can $\crit(j)$ be $<\lambda$-extendible?

**Edit:**
I should point out (reminded by Carlo Von Shnitzel's comments below) that there is a sort of local intertwining of supercompact cardinals and extendible cardinals that may be relevant. See Kanamori's book, p.316-318.

Also, there may be some subtlety here concerning $\Sigma_k$ correctness. Suppose $$j:V_\lambda\prec V_\lambda.$$ I think assuming $V_\lambda\prec_3 V$ (or even $V_\lambda\prec_2 V$) is a strictly stronger assumption. If $\crit(j)$ were extendible, then $V_{\crit(j)}\prec_3 V$. But the embedding assumption also gives us that $V_{\crit^\omega(j)}\prec_3 V$, even though $\crit^\omega(j)=\lambda$ is not itself an extendible cardinal. Similarly if we assume $\crit(j)$ is *actually* supercompact.