A cardinal $\kappa$ is Reinhardt if there is a nontrivial elementary embedding $j:V\to V$ such that $\kappa$ is the critical point of $j$. Kunen proved (using choice) that this large cardinal axiom is inconsistent.
For ordinal $\lambda$, a cardinal $\kappa$ is called $\lambda$-strong if there is a nontrivial elementary embedding $j:V\to M$ for some transitive class $M$ where $V_\lambda\subseteq M$, and such that $\kappa$ is the critical point of $j$. A cardinal $\kappa$ is called strong if it is $\lambda$-strong for all ordinals $\lambda$. As far as I know, strong cardinals continue to be studied with the axiom of choice.
But it appears that all strong cardinals are Reinhardt. Assume $\kappa$ is a strong cardinal. Then there is a nontrivial elementary embedding from $V$ to a transitive proper class $M$ such that for all $\lambda$, $V_\lambda\subseteq M$ (and the critical point of $j$ is $\kappa$). This means that the class $M$ is necessarily $V$. In this case, it looks like all strong cardinals are Reinhardt. Where is the problem with this line of reasoning?
The problem is that there is a quantifier exchange fallacy in deducing that if $\kappa$ is strong, there is a nontrivial elementary embedding from $V$ to a single transitive class $M$ such that $V_\lambda\subseteq M$ for all ordinals $\lambda$. The positions of the "exists $M$" and "for all $\lambda$" quantifications are switched:
In particular, strongness of $\kappa$ gets you a transitive class $M_\lambda$ and a nontrivial elementary embedding $j_\lambda:V\to M_\lambda$ for each ordinal $\lambda$, where $V_\lambda\subseteq M_\lambda$ and $\mathrm{crit}(j)=\kappa$.