i have been asked to solve the quastion , when i tried and i found it better to prove first the lemma (both the lemma and the quastion need to be proven):
let $\sigma \in S_n$ be a permutation from {1,...,n} to {1,...,n} so prove that,
lemma: $\forall x \in ${1,...,n}$ : \sigma (x) \neq x \ so \ \sigma (\sigma(...(\sigma(x))...) [n \ times] = x \ \ , \forall x \in${1,...,n}
now the question is : whats the probability that by choosing arbitrary permutation $\sigma (\sigma(...(\sigma(x))...) [n \ times] = x \ \ , \forall x \in${1,...,n} ?
please try to prove them both but even a solution to the lemma alone will be excepted - thanks for help.
No answer but too much for a comment.
A route to solve the question is the following: find the tuples $\langle a_1,\dots,a_n\rangle$ of nonnegative integers exist with:
Here $a_k$ denotes the number of cycles of length $k$ in the complete disjoint cycle presentation of $\sigma$.
A tuple $\langle a_1,\dots,a_n\rangle$ that satisfies the conditions stands for a conjugacy class that contains elements $\sigma$ that satisfy $\sigma^n=\mathsf{id}$ and the number of elements of the class is: $$\frac{n!}{\prod_{k=1}^nk^{a_k}\cdot a_k!}$$
So if $A$ denotes the collection of these tuples then we find a probability: $$\sum_{\langle a_1,\dots,a_n\rangle\in A}\frac{1}{\prod_{k=1}^nk^{a_k}\cdot a_k!}$$
I hope that there is simpler route.