A cardinal is Reinhardt if $\kappa$ is the critical point of a nontrivial elementary embedding of $V$ to itself, where $V$ is the class of all sets. As Reinhardt cardinals are inconsistent with $\mathrm{ZFC}$, work in $\mathrm{ZF}j$, which is $\mathrm{ZF}$ with replacement and separation for formulas including the function symbol $j$, and an axiom schema stating $j$ is a nontrivial elementary embedding $V\to V$.
Let $(V_\alpha)_{\alpha\in\mathrm{Ord}}$ be the usual von Neumann hierarchy whose union is $V$. For two elementary embeddings $j,k:V\to V$, define the embedding $j\cdot k$, the application of $j$ to $k$, to be $\bigcup_{\alpha\in\mathrm{Ord}}j(k\cap V_\alpha)$. The application operation is familiar from the study of rank-into-rank axioms like $\mathrm I3$, a usual reference is Laver's "On the Algebra of Elementary Embeddings of a Rank Into Itself". It is a usual exercise to verify that if $j$ and $k$ are elementary, then $j\cdot k$ is elementary.
If there exists a nontrivial elementary embedding $j:V\to V$ along with a Reinhardt cardinal $\kappa$, then $j\cdot j$ is also elementary, and $j(\kappa)$ is also Reinhardt as it is the critical point of $j\cdot j$. Similarly, $j\cdot (j\cdot j)$ is elementary, and $j(j(\kappa))$ is Reinhardt. More generally, define the critical sequence of $j$ as usual, by $\kappa_0=\kappa$, $\kappa_{n+1}=j(\kappa_n)$, and $\lambda=\mathrm{sup}\{\kappa_n\mid n<\omega\}$. Additionally define $j^0=j$ and $j^{n+1}=j\cdot j^n$. Each $\kappa_n$ in the critical sequence is Reinhardt, as witnessed by the elementary embedding $j^n$. So if there exists a Reinhardt cardinal, there must exist countably many.
Assume there exists a Reinhardt cardinal, and let $\alpha$ be an arbitrary von Neumann ordinal. Does it follow that there is a Reinhardt cardinal $>\alpha$?
Some possible direction: In section 6 of "I0 and rank-into-rank axioms", Dimonte considers direct systems $(M_\beta,j_{\beta,\gamma})_{\beta<\gamma<\alpha}$ of elementary embeddings (although these embeddings are only from $V_\lambda$ to $V_\lambda$), taking direct limits at limit ordinal steps, but due to the different definition $j^{\alpha+1}=j^\alpha\cdot j^\alpha$, if I am correct the critical point of $j^2$ is no longer $j(j(\kappa))$, but only $\kappa$, so no larger Reinhardt cardinals are produced.
Best Answer
No, if the existence of a Reinhardt is consistent, then it is consistent with a Reinhardt cardinal that the class of inaccessible cardinals is bounded in the ordinals. Indeed, if $j : V\to V$ is a nontrivial elementary embedding and $\lambda$ is the least fixed point of $j$ above its critical point, then let $\delta$ be the least inaccessible above $\lambda$ if there is one -- if not, we're done. Note that $j(\delta) = \delta$, so $(V_\delta,V_{\delta+1})$ is a model of full second-order ZF + there is a Reinhardt cardinal + every inaccessible cardinal is less than $\lambda$.
One interesting question here is whether it is consistent that there are exactly $\omega$-many Reinhardt cardinals (in ordertype). It seems like one might have a chance of showing this is true in the model of NBG where all classes are definable from a single elementary $j : V\to V$.