Given two fixed $n$-cycles in $S_n$, $\alpha$ and $\beta$, I want to find how many unique $\sigma \in S_n$ satisfy
$$\sigma \alpha \sigma^{-1} = \beta$$
In class, we have already learned that conjugation preserves cycle type. I have attacked this problem several times and believe it to be $n$, but I would like to verify if this proof holds and if there is an easier way to do this. My attempt goes like:
- Fix an element $a_1$ in ${1,2,...,n}$
- Represent $\alpha$ as $(a_1, \alpha(a_1), \alpha^2(a_1), ... \alpha^{n-1}(a_1) )$
- Denote $\Sigma$ the set of all permutations $\sigma$ satisfying the above
- Partition $\Sigma$ into $n$ disjoint subsests ${\Sigma_1, \Sigma_2, ... \Sigma_n}$, where $\sigma \in \Sigma_i$ if $\sigma(a_1) = i$
- Construct the permutation $\sigma_i$, which we define:
$$ \sigma_i(a_1)=i \\ \sigma_i(a_i)=\beta^{i-1}(i) $$
Hence the cycle representation of $\sigma_i \alpha \sigma_i^{-1}$ is $$(\sigma_i(a_1), \sigma_i(a_2),...,\sigma_i(a_n))=(i,\beta(i),\beta^2(i),...,\beta^{n-1}(i))$$
The right side is a representation of $\beta$, so we have $\sigma_i \in \Sigma_i \subset \Sigma$
Suppose a given subset $\Sigma_i$ had some other element in it, $\bar{\sigma_i}$. Then $\bar{\sigma_i}$ and $\sigma_i$ must agree on $a_1$ since they both lie in $\Sigma_i$, and they must disagree on some other element $a_l$. We may as well choose $l$ such that it is the minimal such positive index which this occurs- i.e., $\sigma_i$ and $\bar{\sigma_i}$ agree for all $a_i$ for $i < l$. Then the cycle representation of $\bar{\sigma_i} \alpha \bar{\sigma_i}^{-1}$ is
$$(\bar{\sigma_i}(a_1),\bar{\sigma_i}(a_2),...\bar{\sigma_i}(a_n)) = (i,\beta(i),...,\beta^{l-2}(i), \mathbf{\bar{\sigma_i}(a_l)}, \bar{\sigma_i}(a_{l+1}),...\bar{\sigma_i}(a_n)) $$
I have put the $l$th element in boldface to indicate that it is strictly not equal to $\beta^{l-1}(a_l)$ by assumption. Clearly this is a contradiction because this is no longer a representation of the cycle $\beta$. Thus we cannot construct any members of $\Sigma_i$ which differ from the permutation $\sigma_i$. That is, each $\Sigma_i$ is a singleton.
Thus, the collection of permutations $\Sigma$ that we want is partitioned into exactly $n$ singleton sets, and hence $|\Sigma|=n$.
I am not very happy with the proof I have given, as essentially it consists of explicitly constructing all of the permutations and proving their uniqueness, so it seems like the least effective method. It seems very plausible to me that this could be done somehow using group actions, but I'm not sure how. Any help would be appreciated.