How can I show that this permutation equation has 5 solutions:
$\pi^{2013}$ = (1 9) (2 8) (3 7) (4 6) (5)
Since the cycle structure is [2, 2, 2, 2, 1] then the only possible cyclic structure for the solution $\pi$ is the same one since gcd(2, 2013) = 1 and all other multiples of 2 would yield the same result. Now 2013 mod 2 = 1 so $\pi^1$. Isn't that the only solution?
Hint
If $\ a,b,c,d,e,f\ $ are distinct, then $$ \pmatrix{a& b&c&d &e &f}^3=\pmatrix{a&d}\pmatrix{b &e}\pmatrix{c&f}\ , $$ and $\ 2013=3\times671\ .$ Therefore \begin{align} \pmatrix{a& b&c&d &e &f}^{2013}&=\pmatrix{a& b&c&d &e &f}^{3\times671}\\ &=\big(\pmatrix{a&d}\pmatrix{b &e}\pmatrix{c&f}\big)^{671}\\ &=\pmatrix{a&d}\pmatrix{b &e}\pmatrix{c&f}\ . \end{align}