“…Clearly, the automorphism group of F 16n for n ≥ 3 can be computed similarly. If α denotes the rotation of the F 16n through an angle of 90 • around an axis through the midpoints of the front and back faces, then the corresponding permutation is α = (1, 3, 5, 7) (2, 4, 6, 8)(9, 15, 26, 21) (10,16,27,32) (11,17,28,22) (12,18,29,23) (13,19,30,24) (14,20,31,25) (33,45,41,37) (34,46,42,38) (35,47,43,39) (36,48,44,40). Thus the orbit of the subgroup α is 1 α = {1, 3, 5, 7}.…”