Conjugacy class of $(12\cdots n)$ in $S_n$.

Question

Claim: in $S_n$, the conjugacy class of $(12\cdots n)$ is of cardinality $(n-1)!$.

I'm not sure how to prove this. Call this permutation $\gamma$. If $\alpha$ is conjugate to $\gamma$ then there exists $\beta$ such that:

$$\alpha = \beta \gamma \beta^{-1} = (\beta(1) \beta(2) \cdots \beta(n))$$

I guess I have to count how many possibilities are there for $\beta$, and this I'm not sure how to do.

Hint or answer much appreciated.

Answer 1

Your argument shows the conjugates of $\gamma$ are precisely the $n$-cycles $(a_1\,a_2\,\ldots\, a_n)$. Each $n$-cycle can be uniquely written with $a_1=1$ etc.