Let $j:V\to M$ be an elementary embedding. Assume $\xi\geq crit(j)$ is strongly inaccessible, and $\alpha<\xi$. Is it possible that $j(\alpha)\geq\xi$ ?
If yes - does the additional assumption $V_\xi \subset M$ change it?
Note that if $j$ is an ultrapower embedding ($j=j_U$ for some ultrafilter $U$), then the answer is indeed "no" - if $\xi>crit(j)$ is strongly inaccessible then $j(\xi)=\xi$ so $\alpha<\xi$ implies $j(\alpha)<\xi$.
Of course it's possible, with sufficiently large large cardinals.
If $\kappa$ is supercompact and $\xi>\kappa$ is inaccessible, then there is an elementary embedding $j$ such that $j(\kappa)>\xi$, in fact, there is one with $\kappa$ as the critical point. In fact, in that scenario, we can ask the target model to be closed under $\xi$-sequences, which in case $\xi$ is a strong limit, we get $V_\xi\subseteq M$.
Superstrong cardinals are those where $V_{j(\kappa)}\subseteq M$, and those are weaker than supercompact.
And finally, note that while it's true that supercompact embeddings are not ultrapowers like measurable cardinals are, but they are ultrapowers by larger sets and "better" measures: namely $\mathcal P_\kappa(\xi)$ and some fine and normal $\kappa$-complete measure on it.
In fact, just asking that $j(\kappa)$ is arbitrary large can be done with just a single measurable cardinal. We simply pick an embedding (your favorite ultrapower, for example), and iterate it. Gaifman's theorem tells us that the direct limits are well-founded as well, and it is easy to show that the critical points must always strictly increase.
This is also a way of building reasonably canonical inner models. If you start with the smallest measurable cardinal in your universe, and you keep iterating the embedding, then you necessarily kill your measurable cardinals, one at a time. So the result is an inner model which does not have a measurable cardinal in it. And in fact, this is a way of building the Dodd-Jensen core model.