= Solution
A point fixed by $F^4$ satisfies $(2^4-1)x\in\mathbb Z$, so $x=j/15$ for $0\leq j\leq14$. The points with period dividing two are $0,1/3,2/3$, corresponding to $j=0,5,10$. All other twelve points have least period four. Multiplication by two modulo fifteen groups them into the three cycles
$$
\boxed{(1/15,2/15,4/15,8/15),\quad
(3/15,6/15,12/15,9/15),\quad
(7/15,14/15,13/15,11/15).}
$$
This exhausts the requested four-cycles. The displayed order is the dynamical order, and cyclic changes of the starting point describe the same cycle.
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