 Solution

A point fixed by $F^4$ satisfies $(2^4-1)x\in \mathbb{Z}$, so $x=j/15$ for $0\le j\le 14$. 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 cyclesThis exhausts the requested four-cycles. The displayed order is the dynamical order, and cyclic changes of the starting point describe the same cycle.