Conjugacy Classes of subgroups in GL(n,p) What is the conjugacy class of subgroups of order $p$ in $GL(n,p)$? (Are all subgroups of order $p$ conjugate in $GL(n,p)$?)
 A: Every subgroup of order p in GL(n,p) is conjugate to a subgroup generated by a matrix in Frobenius normal form.  Such a matrix is a direct sum of companion matrices of the polynomials (x−1)k where 1 ≤ k ≤ p, and is determined up to conjugacy by the partition of n defined by 1 ≤ k1 ≤ k2 ≤ … ≤ km ≤ p where n = k1 + k2 + … + km and all such partitions other than 1 ≤ 1 ≤ … ≤ 1 define subgroups of order p (k1 = k2 = … = kn = 1 defines the trivial subgroup).
These ki are calculated from the subgroup by considering the dimensions of subspaces fixed by the subgroup (eigenspaces of eigenvalue 1) and the associated generalized eigenspaces (nullspaces of those polynomials (x−1)k), so this is just a more detailed version of @jug's comment.
If we let I(n,p) be the number of conjugacy classes of subgroups of GL(n,p) of order p, then we have the bijective but slightly messy formula:

I(n,p) = number of partitions (disregarding order) of n into positive integers no greater than p.
  In particular, no, GL(n,p) contains non-conjugate subgroups of order p as soon as n ≥ 3 if p is odd, and as soon as n ≥ 4 if p = 2.

For p = 2, one gets the simple formula I(n,2) = floor(n/2).
I believe this is also equal to NrPartitions(n+p,p)−1, the number of partitions of n+p into at most p things (and subtracting one to remove the trivial subgroup).  This should follow by dual Young tableaux or otherwise by general combinatorics.
A: (Same as Jack's, different normal form!)
If $A\in\operatorname{GL}(n,p)$ has order $p$, then $A^p-I=0$. It follows that $A$ satisfies the polynomial $X^p-1$, so the minimal polynomial $m_A$ divides $X^p-1=(X-1)^p$. Since every eigenvalue of $A$ is a root of $m_A$, we conclude that the only eigenvalue of $A$ (in any algebraic closure of $\mathbb F_p$), is $1$.
It follows at once that both the Jordan form $J$ of $A$ and a matrix conjugating $A$ to $J$ are in fact both elements of $\operatorname{GL}(n,p)$. If we want to count the elements of $\{A\in\operatorname{GL}(n,p):A^p=I\}$ up to conjugation in $\operatorname{GL}(n,p)$, then, it is enough to count the conjugacy classes of matrices of $\operatorname{GL}(n,p)$ in Jordan canonical form with minimal polynomial dividing $(X-1)^p$. 
This is a pretty simple excercise. The answer is: the number of partitions of $n$ with parts not larger than $p$.
Hmm, this is counting conjugacy classes of matrices and not of subgroups...*
