Solution
= Solution
An itinerary fixed by the eighth shift has least period dividing $8$. Those with period dividing $4$ are exactly the itineraries fixed by the fourth shift. Thus the primitive length-eight itineraries form
$$
\operatorname{tr}(A^8)-\operatorname{tr}(A^4)=47-7=40
$$
pointed words. Each cycle has eight cyclic shifts, so they give
$$
\boxed{40/8=5}
$$
distinct $8$-cycles. Therefore at least five such cycles are forced.