Note that this separates the flies into two groups of four. Each fly in a group can move to any other vertex that a fly in the same group originally occupies. So, this turns into finding the number of derangements of {1,2,3,4} so that none of the numbers are located on their correct position, and then this number is squared since there are two independent groups. The number of derangements of four elements is 9 (Can be tested with simple trial and error, as there are only 24 cases to consider), so the number of possible ways for the flies to move is $9^2 = \boxed{81}$.

Separate into two tetrahedra. Then the flies can either have 2 cycles of 2 or one cycle of length 4. For 2 cycles it's just $\tfrac{\binom{4}2}2=3$. For 1 cycle notice that two out of six edges will not be traversed, and they are opposite of each other. So again we get $\tfrac{\binom{4}2}2=3$, but we multiply by $2$ to account for chirality. So we get $9$ which squared is $\boxed {81}$.