I don't know, what is meant by "identical permutation", but the following is likely to answer your question. Let $s(\chi)$ denote the Frobenius-Schur indicator of the irreducible character $\chi$. Thus, $s(\chi)=1$ if the representation of $\chi$ can be realised over $\mathbb{R}$, $s(\chi)=-1$ if $\chi$ is real-valued but the corresponding representation is not realisable over $\mathbb{R}$ and $s(\chi)=0$ if $\chi$ is not real-valued. Then, the number of square roots of an element $g$ in any group is equal to
$$\sum_\chi s(\chi)\chi(g),$$
where the sum runs over all irreducible characters of the group.

Of course, in $S_n$, all Frobenius-Schur indicators are 1, so the number of square roots of $x_0$ is just $\sum_\chi \chi(x_0)$. Since the representation theory of $S_n$ is completely understood, I believe that one should be able to give the maximum number of square roots of an element in closed form as a function of $n$ (if that's what you are interested in, which was not clear from the question).

I hope, this is on topic, but I am happy to edit or delete as the question is further clarified.

Edit: Just saw the clarification. This proves that the maximal number of solutions is indeed attained by $x_0 = 1$, since each character value attains its maximum there. This generalises immediately to any group, for which any representation is either realisable over $\mathbb{R}$ or has non-real character. Other cases would require more thought, but this is likely the right way to go about it.