Q:

Find all k, such that $A_{k}= \left \{\sigma ^{k}|\sigma \in S_{n} \right \}$ is a subgroup of $S_{n}$

Find all odd k, such that $A_{k}= \left \{\sigma  ^{k}|\sigma \in S_{n}  \right \}$ is a subgroup of $S_{n}$

A:

If $n\geq 5$ and $k$ is odd, then the set $\{x^k~|~x\in S_n\}$ is a subgroup of $S_n$ if an only if $g.c.d.(n!,k)=1$.
Here we use the classical finite group theoretical fact that the map $x\to x^k$ is a bijection of a finite group $G$ if and only if $g.c.d.(|G|,k)=1$.
If $g.c.d.(n!,k)=1$, then the map $x\to x^k$ is a bijection of $S_n$ and therefore $\{x^k~|~x\in S_n\}=S_n$ is a subgroup.
Conversely, suppose that $\{x^k~|~x\in S_n\}$ is a subgroup. Since $y^{-1}x^ky=(y^{-1}xy)^k$, then this subgroup is normal. For $n\geq 5$ the group $S_n$ has only $3$ normal subgroups, namely $S_n, A_n,1$.
Since $k$ is odd, for an odd permutation $x$ the permutation $x^k$ is also odd. Therefore $\{x^k~|~x\in S_n\}\neq A_n, 1$. Therefore $\{x^k~|~x\in S_n\}=S_n$ and the map $x\to x^k$ is a bijection of $S_n$, which implies $g.c.d.(n!,k)=1$.
If $n\geq 6$ and $k$ is even, then the set $\{x^k~|~x\in S_n\}$ is a subgroup of $S_n$ if and only if $k$ is divisible by $l.c.m.(2,3,\dots,n)$.
Indeed, if $k$ is divisible by by $l.c.m.(2,3,\dots,n)$, the for all $x\in S_n$, we have $x^k=1$, therefore $\{x^k~|~x\in S_n\}=1$ is a subgroup.
If $k$ is not divisible by $l.c.m.(2,3,\dots,n)$, then the set $\{x^k~|~x\in S_n\}$ is not empty. 
Suppose that $\{x^k~|~x\in S_n\}$ is a subgroup of $S_n$. This subgroup is non-empty normal subgroup which contains only even permutations. Therefore $\{x^k~|~x\in S_n\}=A_n$.
Denote by $H=\{x^2|x\in S_n\}$. This set is obviously a subset of $A_n$. 
If $x$ is even cycle, then $x^2$ is a product of two cycles of the same length. 
If $x$ is odd cycle, then $x^2$ is again an odd cycle.
Therefore the element $x\in S_n$ is a square if and only if for every even $r$ the number of cycles of the length $r$ in $x$ is even. Since the element $y=(12)(3456)$ does not satisfy this condition, it is not a square. Therefore $y$ belongs to $A_n\setminus H$, and $H\neq A_n$.
Finally  $\{x^k~|~x\in S_n\}=\{x^{k/2}~|~x\in H\}$ is a proper subset of $A_n$, i.e. not equal to $A_n$. We have contradiction.
Oops, I didn't catch that the question is stated for $k$ odd only.

