So, a Reinhardt cardinal does indeed satisfy I1, I-1, etc by construction.

Huh; I must have had a brain fade there. If $\lambda < \kappa$, then any well-ordering of $V_\lambda$ is in $V_\kappa$, so its length must be fixed by j and so less than $\kappa$. Thus the Hartogs cardinal of $\lambda$ must be $\kappa$ or below; but it is fixed by j as well, so it is less than $\kappa$. It follows that $2^\lambda$ is incomparable to or less than $\kappa$.

Oh, but we don't necessarily know that $\delta$ is strong limit; if $\lambda < \kappa$, then $j(2^\lambda) = 2^j(\lambda) = 2^\lambda$, but this doesn't preclude the possibility that $2^\lambda > \delta$, nor that $2^\lambda$ is an alternate least fixed point of $j$ beyond the $\kappa_n$. So I guess we only get a partial version of the rank-into-rank axioms. But I think Aspero's result implies that there is some $\mu$ which has, for each $\eta$, $i_\eta:V_{\nu+\eta} \to V_{\nu+\eta}$ (\nu the LFP for i_\eta), and $i_\eta(\bar{\kappa})$ can be chosen greater than the Hartog cardinal for $\mu$.

Yeah, that still works, since $j(rank(S)) = rank(j(S))$ for any set $S$.

Hmmm. It may not be so easy after all. I was figuring that any generic well-ordering of $V_\delta$ would do; but it also needs, at least, a property ensuring consistency with $j$. In fact, the failure of Choice on $V_{\delta+2}$ should transfer to $V_{\kappa+2}$, and hence to a stationary sequence of smaller sets as well. So a better approach would be to use forcing over HOD to get a restriction of $j$ into the extended model.