How many conjugacy classes of elements of order 7 are there in $GL(6, \mathbb{F}_2)$? 
How many conjugacy classes of elements of order 7 are there in $GL(6, \mathbb{F}_2)$?

I've been thinking the following, every element of order 7 must satisfy the next equation:
$$x^7-1=0$$
We can express $$x^7-1=(x-1)(x^3+x+1)(x^3+x^2+1)$$
My intuition is that there are two conjugacy classes determined by the roots of the polynomials $(x^3+x+1)$ and $(x^3+x^2+1)$, i.e, $Q=\{A\in GL(6, \mathbb{F}_2) : A^3+ A + I=0\}$ and $P=\{A\in GL(6, \mathbb{F}_2): A^3+A^2+I=0\}$ are the two conjugacy classes with elements of order 7.
Also, I know that if some element $A\in GL(6, \mathbb{F}_2)$ is in $P$ (or in $Q$ respectively) then, every element in the conjugacy class of $A$ is in $P$ (or $Q$).
Is my intuition right?
Is there any other idea to solve this?
 A: You have correctly identified two of the five conjugacy classes.
$Q$ consists of those 6x6 matrices conjugate to two block diagonal pieces such as:
$$Q = \{ A^{-1}Q_6A : A \in \operatorname{GL}(6,\mathbb{F}_2) \} \qquad Q_6 = \begin{bmatrix} Q_3 & 0 \\0 & Q_3 \end{bmatrix} \qquad Q_3 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix}$$
$P$ consists of those 6x6 matrices conjugate to two block diagonal pieces such as:
$$P = \{ A^{-1}P_6A : A \in \operatorname{GL}(6,\mathbb{F}_2) \} \qquad P_6 = \begin{bmatrix} P_3 & 0 \\0 & P_3 \end{bmatrix} \qquad P_3 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & 0 & -1 \end{bmatrix}$$
However, you can mix and match, for example
$$R = \{A^{-1}R_6A : A \in \operatorname{GL}(6,\mathbb{F}_2) \} \qquad R_6 = \begin{bmatrix} P_3 & 0 \\0 & Q_3 \end{bmatrix}$$
And you haven't used the other irreducible factor yet: $x-1$ gives you a 1x1 identity matrix. You can't ONLY use this (otherwise you just get the identity matrix), but you can use three of them and then 1 of $Q_3$ or $P_3$.
This gives two more classes:
$$S = \{A^{-1}S_6A : A \in \operatorname{GL}(6,\mathbb{F}_2) \} \qquad S_6 = \begin{bmatrix} I_3 & 0 \\0 & Q_3 \end{bmatrix}$$
$$T = \{A^{-1}T_6A : A \in \operatorname{GL}(6,\mathbb{F}_2) \} \qquad T_6 = \begin{bmatrix} I_3 & 0 \\0 & P_3 \end{bmatrix}$$
The combinatorial structure assigns to each irreducible factor $f$ a multiplicity $m_f$ so that $\displaystyle\sum_{f \in \mathcal{F}} m_f \deg(f) = 6$ where $\mathcal{F}=\{(x-1), (x^3+x+1), (x^3+x^2+1) \}$ are the monic irreducible factors of $x^7-1$.
So we have:
$$\begin{array}{rccc}
& m_{x-1} & m_{x^3+x+1} & m_{x^3+x^2+1} \\ \hline
Q & 0 & 2 & 0 \\
P & 0 & 0 & 2 \\
R & 0 & 1 & 1 \\
S & 3 & 1 & 0 \\
T & 3 & 0 & 1 \\
\end{array}$$
Along with the elements of order dividing 7 properly, the identity:
$$\begin{array}{rccc}
& m_{x-1} & m_{x^3+x+1} & m_{x^3+x^2+1} \\ \hline
I & 6 & 0 & 0
\end{array}$$
