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Is the proof of existence of Reinhardt (and higher) cardinals violating Choice dependent on Extensionality in an essential manner?

What I mean is if we work in $\sf ZFA$ would it be possible to have a model that satisfy existence of Reinhardt cardinals and yet satisfy choice?

I ask this question because I saw that $\sf NF$ for example is incompatible with choice but just weakening Extensionality as to allow existence of Ur-elements resulted in compatibility with choice. So can a similar situation be raised here?

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    $\begingroup$ Since being a pure set is a first-order property, any elementary embedding of the universe is also an elementary embedding of the kernel, which is a model of ZFC, if you assume choice. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 12, 2022 at 18:22

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