What is the group generated by the conjugacy class containing $(12\ldots n)$ in $S_n$? This question is motivated by this answer to a question about groups generated by conjugacy classes.
Let $n \geq 1$ and $S_n$ be the symmetric group on $\{1,2,\ldots,n\}$.  Define the n-cycle $\alpha=\pmatrix{1&2&\cdots &n}$, and let $\operatorname{Cl}(\alpha)=\{\beta \alpha \beta^{-1}:\beta \in S_n\}$ denote the conjugacy class of $\alpha$.

Question: What is the group $\langle \operatorname{Cl}(\alpha) \rangle$?

Observation 1: If $n$ is odd and $n \geq 3$, then $\alpha$ and its conjugates are even permutations (since they all have the same cycle structure), so $\langle \operatorname{Cl}(\alpha) \rangle$ contains only even permutations, and thus is a subgroup of the alternating group $A_n < S_n$.
Observation 2: Judging from some computations in GAP, it looks like $\langle \operatorname{Cl}(\alpha) \rangle=A_n$ for odd $n \geq 3$ and $\langle \operatorname{Cl}(\alpha) \rangle=S_n$ for even $n \geq 2$.

Your conjecture is right. First, $\text{CL}(\alpha)$ is precisely all the $n$-cycles, from the simple fact that if $\beta \in S_n$, then 
$$
\beta(12\dots n)\beta^{-1} = (\beta(1) \beta(2) \dots \beta(n)).
$$
Suppose $n \ge 5$ is odd. Then the $n$-cycle is an even permutation, so we can write it as a product of even permutations. The product $(abc d \dots n) (n \dots d bc a) = (bdc)$ gives us an arbitrary $3$-cycle, so that the $n$-cycles generate $A_n$. But since every $n$-cycle is even this is all it can generate. (For $n=3$, the $n$-cycles are $3$-cycles so they obviously generate $A_3$).
Suppose $n \ge 4$ is even. Then the $n$-cycle is an odd permutation. Taking the product $(abc \dots n)(n \dots cab) = (acb)$ gives us again $A_n$, but then $A_n$ is normal in $S_n$ (it has index $2$), so that any odd permutation will be generated by any $n$-cycle multiplied by some unique element of $A_n$. Therefore for $n$ even one gets $S_n$.
Hope that helps,
