Suppose we have an NP-complete language $L_1$ and a NEXP-complete language $L_2$. For any deterministic exptime machine $M_1$ with oracle access $M_1^{L_1}$, is it possible to find a deterministic exptime oracle machine $M_2$ with access $M_2^{L_2}$ such that (a) $M_2$ may only make poly(n) length queries to $L_2$ (b) $M_2^{L_2}$ accepts iff $M_1^{L_1}$ accepts? (Note $M_1$ is capable of making exp(n) length queries to $L_1$ as it is an exponential time TM).
If the above is not true for a particular $L_2$, is it possible to find an $M_2$ and an $L_2\in$NEXP such that the above is true?
Obviously, there is always a polytime reduction from $L_1$ to $L_2$ as $L_2$ is NEXP-hard and $NP\subseteq NEXP$. However if the queries to $L_1$ have $exp(n)$ length, then under the polytime reduction the corresponding $L_2$ instances will now also have $exp(n)$ length. Hence, if $M_2$ is restricted to only $poly(n)$ length queries it's not clear $M_2^{L_2}$ can always make the necessary queries.
It does not seem unreasonable that given an $(M_1, L_1)$ pair, that $M_2^{L_2}$ can simulate $M_1^{L_1}$ and return the same output. If we have an NP language with $exp(n)$ input, a non-deterministic TM of runtime $O(exp(n))$ is capable of solving it. A NEXP machine also has an $exp(n)$ runtime but on an input of length $poly(n)$ and so might be capable of solving the exponential length NP instance.
Edit: I suppose this boils down to the question, if $EXP_{poly}^A$ is an exponential time oracle machine which is only allowed to make polynomial length queries to $A$, does the following hold: $EXP_{poly}^{NEXP} = EXP^{NP}$? The containment $EXP_{poly}^{NEXP} \subseteq EXP^{NP}$ seems to be straightforward to prove.