Finding representatives of conjugacy classes in $\text{Mat}_2(\mathbb{Q})$ Let $S= \{A \in \text{Mat}_2(\mathbb{Q}) : A^6 = I$ and $A^n \ne I$ for any $0 < n < 6\}$.
I wish to describe the orbits of each of the element in $S$ with respect to conjugation by $GL_2(\mathbb{Q})$ on  $\text{Mat}_2(\mathbb{Q})$.
I can't see how to start. Hints are much appreciated!
 A: Let $A\in S$.  From the condition that $A^6=I$ and $A^n\neq I$ for $0<n<6$, we conclude that the characteristic polynomial of $A$ must be $x^2-x+1$.  We claim that $A$ is conjugate to 
$$X:=\begin{bmatrix}1&1\\ -1&0\end{bmatrix}\,.$$ 
That is, $S$ is a single orbit with representative $X$.

 Suppose that $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\,,$$ where $a,b,c,d\in\mathbb{Q}$ are such that $a+d=1$ and $ad-bc=1$.  Take $$V:=\begin{bmatrix}1&1\\c-d&-a+b\end{bmatrix}\,.$$ Then, $\det(V)=-a+b-c+d$. First, we claim that $\det(V)\neq 0$.  Suppose contrary that $\det(V)=0$. That is,  $$a+c=b+d\,.$$  Since $d=1-a$, we have $$c=b-2a+1\,.$$  As $ad-bc=1$, we get $a(1-a)-b(b-2a+1)=1$, or $$(a-b)^2-(a-b)+1=0\,;$$ nonetheless, the equation $t^2-t+1=0$ has no solution $t\in\mathbb{Q}$.  This is a contradiction, and the claim is proven. Note that $$X\,V=\begin{bmatrix}1&1\\-1&0\end{bmatrix}\,\begin{bmatrix}1&1\\c-d&-a+b\end{bmatrix}=\begin{bmatrix}1+c-d&1-a+b\\-1&-1\end{bmatrix}$$ and $$V\,A=\begin{bmatrix}1&1\\c-d& -a+b\end{bmatrix}\,\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a+c&b+d\\-ad+bc&-ad+bc\end{bmatrix}\,.$$ Because $a+d=1$ and $ad-bc=1$, it can be seen immediately that $X\,V=V\,A$, whence $A$ is conjugate to $X$.

A: Every $A \in \operatorname{Mat}_2(\Bbb Q)$ has a conjugate element $PAP^{-1}$ (with $P \in GL_2(\Bbb Q)$) which is in its Frobenius normal form.  Since $S$ is closed under conjugation, we can represent each orbit by enumerating the elements of $S$ which have this normal form.
Any element of $S$ will have a minimal polynomial which divides $x^6 - 1$. Over $\Bbb Q$, the polynomial $x^6 - 1$ factors into the product
$$
p(x) = x^6 - 1 = (x-1)(x+1)(x^2 - x + 1)(x^2 + x + 1)
$$
We eliminate any matrices whose minimal polynomial divides $x-1$, $x^2 - 1$, or $x^3 - 1$.  The only monic polynomial over $\Bbb Q$ which divides $p$ but none of these smaller polynomials is $q(x) = x^2 - x + 1$.  Thus, $q$ must be the minimal polynomial of our matrix in $S$.
Conclude that $S$ contains exactly one orbit, which belongs to the representative
$$
A = \pmatrix{0&-1\\1&1}
$$
A: Let $A$ be a matrix in $S$. Then its minimal polynomial divides the annihilator polynomial $x^6-1=(x^2 + x + 1)(x^2 - x + 1)(x + 1)(x - 1)$. If $A$ has a rational eigenvalue, than the other one is also rational, so both eigenvalues are among $\pm 1$, so $A^2=I$, contradiction. So $S$ has some non-rational eigenvalue. This must be a root of one of the factors $ (x^2 \pm x + 1)$, so the other eigenvalue is the Galois / complex conjugate. We exclude than the factor dividing $x^3-1$, because then $A^3$ would be $I$, contradiction. So the minimal polynomial of $A$ is $x^2-x+1$, thus we have a simpler characterization of $S$, 
$$
S=\left\{\ A=\begin{bmatrix}a& b\\ c& d\end{bmatrix}\ :\
\operatorname{Trace}(A)=a+d=1\ ,\ \det A= ad-bc=1\ \right\}
\ .
$$
Consider now two matrices $A,B$ from $S$ with entries $a,b,c,d$, and respectively $s,t,u,v$,
$$
A =
\begin{bmatrix}a& b\\ c& d\end{bmatrix}
\ ,
\qquad
B=\begin{bmatrix}s & t \\ u& v\end{bmatrix}
\ .
$$
We want (and show how) to conjugate them in each other. Of course, $bc\ne 0$, so a conjugation as follows
$$
\begin{bmatrix}1& n\\ 0& 1\end{bmatrix}
\begin{bmatrix}a& b\\ c& d\end{bmatrix}
\begin{bmatrix}1& -n\\ 0& 1\end{bmatrix}
=
\begin{bmatrix}a+nc & *\\ *& *\end{bmatrix}
\begin{bmatrix}1& -n\\ 0& 1\end{bmatrix}
=
\begin{bmatrix} a+nc & *\\ *& *\end{bmatrix}
$$
can be found to reduce the situation to one with $a=s$. The trace condition implies $d=v$. Now $bc=tu$ (not zero) from the determinant condition, and a conjugation with a diagonal matrix (with diagonal $1$ and $t/b$) solves the problem.
