Permutation cube identity criterion ($f^2=f^{-1}⟺f^3=\mathrm{id}$)

For a bijection $f$, composing $f^2=f^{-1}$ with $f$ gives $f^3=\mathrm{id}$, and composing the latter with $f^{-1}$ gives the converse. Every orbit therefore has size one or three: the permutation consists of fixed points and disjoint three-cycles. This applies to an infinite underlying set as well as a finite one, since each individual orbit has at most three elements.