TITLE: If permutation matrices are conjugate in $\operatorname{GL}(n,\mathbb{F})$ are the corresponding permutations conjugate in the symmetric group?
QUESTION [16 upvotes]: There is a standard embedding of the symmetric group $S_n$ into $\operatorname{GL}(n,\mathbb{F})$ (for any field $\mathbb{F}$) that sends each permutation in $S_n$ to the corresponding permutation matrix.  As this is a group homomorphism, certainly, conjugate permutations map to conjugate matrices.  What about the converse?  Suppose that $A$ and $B$ are similar permutation matrices in $\operatorname{GL}(n,\mathbb{F})$ coming from permutations $\alpha$ and $\beta$ in $S_n$.  Does it follow that $\alpha$ and $\beta$ are already conjugate in $S_n$?  (I thought that I had a proof, but I noticed that the argument seems to fail in positive characteristic.)

REPLY [7 votes]: Note: I've edited this proof to replace the induction step with an appeal to the Möbius inversion formula.
Note: I've added a shorter proof in the case of characteristic zero.
This is true in any characteristic. $A$ and $B$ are similar if and only if they define isomorphic $F[X]$-module structures on $F^n$ via $Xv = Av$ and $Xv = Bv$, respectively.
If the cycles of $\alpha$ have lengths $n_1, \dots, n_s$, then the canonical $F[X]$-module structure is isomorphic to $M = \oplus_i F[X]/(X^{n_i} - 1)$. 
I will assume first that the characteristic is $p > 0$. Write $n_i = r_i p^{k_i}$ with $r_i$ prime to $p$. Then $M = \oplus_i F[X]/(X^{r_i} - 1)^{p^{k_i}}$. We want to check that the numbers $r_i$ and $k_i$ are well-determined by $M$ up to permutation. Let $h_k(r)$ denote the number of times the factor $(X^r - 1)^{p^k}$ appears in the sum for $M$.
For any number $r$ prime to $p$, write $\phi_r$ for the $r$th cyclotomic polynomial. The polynomials $\phi_r$ are pairwise relatively prime, and $X^r - 1 = \prod_{d|r} \phi_d$.
We have $M = \oplus_r M_r$ where $M_r$ is the submodule of $M$ annihilated by some power of $\phi_r$, and $M_r = \bigoplus_{i \text{ such that } r|r_i} F[X]/(\phi_r^{p^{k_i}})$. For a fixed $r$, the powers appearing in this decomposition are well determined by $M_r$ (and hence $M$), by the uniqueness part of the structure theorem for finitely generated modules over a PID. 
If we let $f_k(r)$ be the number of times that $\phi_r^{p^k}$ appears in the decomposition of $M_r$, then $f_k(r)$ is well determined by $M$ and we have $f_k(r) = \sum_{q \text{ such that } r|q} h_k(q)$. To finish the proof, it will be enough to show that the function $f_k = f$ uniquely determines the function $h_k= h$.
If we let $N$ be a common multiple of all numbers $q$ for which $h(q) \ne 0$ (or equivalently, for which $f(q) \ne 0$), and we write $f'(t) = f(N/t)$ and $h'(t) = h(N/t)$, then we have $f'(u) = \sum_{t, \, t|u} h'(t)$. It follows from the Möbius inversion formula that $h'$ is determined by $f'$, hence $h$ by $f$.
If the characteristic is zero, argue the same way, but replacing all powers $p^{k_i}$ with $1$.
Note that the argument using the characteristic polynomial fails in positive characteristic $p$, since for example $X^2 - 1 = (X-1)^2$ when $p = 2$.
ALTERNATIVE PROOF IN CHARACTERISTIC ZERO:
Let $h(c)$ denote the number of cycles of length $c$ in a decomposition of $\alpha$. Note that $\operatorname{tr} A$ is the number of fixed points of the permutation $\alpha$. Thus if we let $f(n) = \operatorname{tr} A^n$, then $f(n)$ is also the number of fixed points of the permutation $\alpha^n$, namely
$$f(n) = \sum_{c|n} ch(c).$$
It follows from the Möbius inversion formula that the function $h(c)$ is entirely determined  by $f(n)$, hence by $A$.