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**Swlabr** The easy way: Are you familiar with the orbit-stabiliser theorem? If so, can you think up a way of using this? Which elements commute with your n-cycle?

The harder way: Alternatively, note that when n is odd, conjugation of a long-cycle by (12) must be "undone" by (12). There is no way to get from $\displaystyle (12)\sigma(12)$ without conjugation by (12) when $\displaystyle n$ is odd.

(When dealing with these kinds of problems it is perhaps helpful to remember that $\displaystyle (a \text{ } b \ldots r)^{\sigma} = (a \sigma \text{ } b\sigma \ldots r \sigma)$ - every coefficient is acted on individually by the permutation you are conjugating by).