= Solution
Points whose period divides two solve
$$
(2^2-1)x\in\mathbb Z,
$$
so, after removing the fixed point zero, the unique two-cycle is
$$
\boxed{\frac13\longmapsto\frac23\longmapsto\frac13}.
$$
Points whose period divides four are $j/15$ for $j=0,\ldots,14$. Removing the three points fixed by $F^2$, namely $0,5/15,10/15$, leaves twelve points of exact period four. They form the three orbits
$$
\boxed{
\frac1{15}\to\frac2{15}\to\frac4{15}\to\frac8{15}\to\frac1{15}},
$$
$$
\boxed{
\frac15\to\frac25\to\frac45\to\frac35\to\frac15},
$$
and
$$
\boxed{
\frac7{15}\to\frac{14}{15}\to\frac{13}{15}\to\frac{11}{15}\to\frac7{15}}.
$$
Starting at another point on one displayed orbit gives the corresponding cyclic rotation.
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