Doe the "hacker" get to see all the contents? Or only 50 (half?)

If the hacker needs to break the longest permutation in half (consider the simple permutation x->(x+1), they could ... swap box 0 with box 50, so box 0 points to 51 (->52 .. -> 99 -> 0) and box 50 points to box 1. That makes 2 cycles of size 50, but requires inspecting 50+1 boxes