Is it possible to find $\gamma \in \Bbb S_{10}$ based on given $\gamma^2$?

Question

Given the permutation

$$\gamma^2=\begin{bmatrix}1&2&3&4&5&6&7&8&9&0\\1&4&5&7&6&9&8&0&3&2\end{bmatrix},$$

its disjoint cycle being $(2,4,7,8,0)(3,5,6,9)$. Is it possible to find the corresponding $\gamma$, where $\gamma \in \Bbb S_{10}$?

Answer 1

A permutation is a square if and only if, in its disjoint cycle decomposition, the number of cycles of the same even length is even.

In fact if $\gamma$ has in its decomposition an odd cycle of length $2n+1$, $(j_1,j_2,\dots ,j_{2n+1})$, then $\gamma^2$ transforms this cycle into the cycle $$(j_1,j_3,\dots ,j_{2n+1},j_2,j_4,\dots,j_{2n})$$ which has the same odd length $2n+1$.

On the other hand if $\gamma$ has in its decomposition an even cycle of length $2n$, $(j_1,j_2,\dots ,j_{2n})$, then $\gamma^2$ transforms this cycle into TWO disjoint cycles of the same length $n$, $$(j_1,j_3,\dots ,j_{2n-1})(j_2,j_4,\dots,j_{2n}).$$

Here $(1)(2,4,7,8,0)(3,5,6,9)$ has only one cycle of length $4$ so it can not be the square of a permutation in $\Bbb S_{10}$.

Bonus questions. Are you able to find a square root of $(1,3,5,6,9)(2,4,7,8,0)$? Is it unique?