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 $\lambda$.

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 $j : V\to V$.