Inversion of an odd cycle in an alternating group ($g{\sim }_{A_n}{g}^{-1}⟺n\equiv 1(\mathrm{mod}4)(n\text{ odd})$)

For an odd $n$-cycle $g$, an inverting permutation fixes one point and swaps $(n-1)/2$ pairs, so its sign is $(-1{)}^{(n-1)/2}$. Its other inverting permutations differ by elements of $C_{S_n}(g)=⟨g⟩$, all even. Thus inversion is possible inside $A_n$ exactly when that sign is positive. In particular $A_n$ is not an ambivalent group when $n\equiv 3(\mathrm{mod}4)$.