Permutation group equation: $x^{20} = \sigma$

Well, cycles of odd length are even permutations, and cycles of even length are odd permutations. Here we have three cycles of even length and one cycle of odd length, and so $\operatorname{sign}(\sigma)=(-1)^3\cdot1=-1$. On the other hand, $\operatorname{sign}(x^{20})=(\operatorname{sign}(x))^{20}=1$, and so the equation has no solutions at all.