$\mathsf{EXP}$ vs $\oplus\mathsf{EXP}$ In our recent work, we resolve a computational problem which arose in combinatorial context, under assumption that $\mathsf{EXP} \ne \mathsf{\oplus{}EXP}$, where $\mathsf{\oplus{}EXP}$ is the $\mathsf{EXP}$-version of $\mathsf{\oplus{}P}$. The only paper on $\mathsf{\oplus{}EXP}$ that we found was the Beigel-Buhrman-Fortnow 1998 paper that is cited  on Complexity Zoo. We understand that we can take parity versions of $\mathsf{NEXP}$-complete problems (see this question), but perhaps many of them are in fact not complete in $\mathsf{\oplus{}EXP}$.  

QUESTION: Are there complexity reasons to believe that $\mathsf{EXP} \ne \mathsf{\oplus{}EXP}$?  Are there natural combinatorial problems that are complete in $\mathsf{\oplus{}EXP}$?  Are there some references we might be missing?
In terms of complexity reasons (rather than complete problems): The Hartmanis-Immerman-Sewelson Theorem should also work in this context, namely: $\mathsf{EXP} \neq \oplus \mathsf{EXP}$ iff there is a polynomially sparse set in $\oplus \mathsf{P} \backslash \mathsf{P}$. Given how far apart we think $\mathsf{P}$ and $\oplus \mathsf{P}$ are - e.g. Toda showed that $\mathsf{PH} \subseteq \mathsf{BPP}^{\oplus \mathsf{P}}$ - it would be quite surprising if there were no sparse sets in their difference. 

More directly, if there were no sparse sets in their difference, it would say that for every $\mathsf{NP}$ verifier, if the number of strings of length $n$ with an odd number of witnesses is bounded by $n^{O(1)}$, then the problem [of telling whether there is an odd number of witnesses] must be in $\mathsf{P}$. This seems like quite a striking and unlikely fact.