 Perfect out-shuffle

Interleave the two equal halves of an even finite list, placing the first half's next entry before the second half's next entry. With zero-based positions and length $n>2$, position $i<n-1$ moves to $2i\mathrm{mod}(n-1)$, and the last position stays fixed. For an index in the first half the new position is $2i$; in the second it is $2i-(n-1)$, proving the formula. The least positive period of the whole permutation is the multiplicative order of $2$ modulo $n-1$: equality of the multiplier to one fixes every position, and fixing position one requires that equality. In particular, a 52-entry list has period eight. The empty and two-entry cases are identity permutations with least positive period one.