TITLE: What are the current views on consistency of Reinhardt cardinals without AC?
QUESTION [16 upvotes]: It's well known that Reinhardt cardinals are inconsistent, provided that we have access to axiom of choice, but, as far as I know, we are clueless about this when we don't assume choice. For me, the fact that Reihardt cardinals are inconsistent with choice already makes them, in a way, "wrong", that is, I'd believe personally that Reinhardt cardinals cannot exist in $V$ anyways. However, I have never seen any claims like this. So my question is:

Is it believed that Reinhardt cardinals can be consistent without choice? Are there any solid arguments for/against them? (other than "we haven't found them inconsistent, so they should be consistent")

Thanks in advance.
Note: I added soft-question tag because this question is more about philosophical views. Feel free to remove it if you think otherwise.

REPLY [15 votes]: The following theorems of Woodin may be related:
Theorem. ($ZF$) Assume that $ZFC$ proves the $HOD$ Conjecture. Suppose $\delta$ is an extendible cardinal. Then for all $ \lambda>\delta$  there is no non-trivial
elementary embedding $j : V_{\lambda+2}\to V_{\lambda+2}$ 
Thus (assuming that $ZFC$ proves the $HOD$ Conjecture) one nearly has a
proof of Kunen's inconsistency theorem without using the Axiom of Choice.
Theorem. Assume $ZF$+ there exists a Reinhardt cardinal + there exists a proper class of supercompact cardinals is consistent. Then there exists a genric extension of the universe which satisfies $ZF$ + the axiom of choice + there exists a proper class of supercompact cardinals, and such that in it Woodin's $HOD$ conjecture fails.
As far as I know, Woodin believes the $HOD$ conjecture is true (at least the current methods can not be used to solve the problem), so by the above theorem we may expect to show that at least   "$ZF$+ there exists a Reinhardt cardinal + there exists a proper class of supercompact cardinals" is not consistent.