Representation of Quaternion group in $GL(2,3)$ I am working with the representation of the quaternion group in $GL(2,3)$ generated by $A=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}, B=\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}, C=\begin{pmatrix} -1 & 1\\ 1 & 1 \end{pmatrix}$.  
Now $C=PBP^{-1}$ where $P= \begin{pmatrix} 0 & 1\\ 1& 0 \end{pmatrix}$.  I am trying to show that $A$ and $B$ are similar by an element in $GL(2,3)$. 
They are obviously similar in $GL(2,9)$ because they are both diagonalizable with the same eigenvalues:  Their characteristic polynomials are both $x^2+1$, and since its irreducible over $F_3$ and $F_3$ is perfect, the roots are distinct.   
I feel that because of the relations $A=CB, B=AC, C=BA$,    $A$, $B$, $C$ should all "be on the same footing": $B$ and $C$ similar by P should mean that $A,B$ are similar too, without me having to worry about whether the similarity matrix has entries in $F_3$.  After all I should be able to use jacobi's identity and the matrix P to achieve this.  It has not been working out for some time!
edit:
another thing I am trying -
Probably the best way to do this is to look at all the embeddings of $Q_8 \hookrightarrow GL(2,3)$.  
 A: It's a general fact that if two $n \times n$ matrices with entries in a field $F$ are similar over a field $K$ containing $F,$ they are already similar over $F.$ This is a consequence of the theory of the rational canonical form (which reduces to the Jordan normal form when the field is algebraically closed). What happens in this example is a rather more straightforward than usual of the general theory. Both $A$ and $B$ have minimal (and characteristic) polynomial $x^{2}+1.$ Now when $M$ is any matrix with this minimum polynomial, we can find a vector $v$ such that $\{v,Mv\}$ is linearly independent, so a basis. With respect to the basis $\{v,Mv \}$ we see that (the linear transformation represented by $M$ in the old basis) has matrix $\left(\begin{array}{clcr} 0 & -1\\1& 0\end{array} \right)$. Hence $TMT^{-1} =\left(\begin{array}{clcr} 0 & -1\\1& 0\end{array} \right)$ for some $T \in {\rm GL}(2,3),$ where $T$ is a change of basis matrix. This shows that any $3 \times 3$ matrix with entries in ${\rm GF}(3)$ and minimum polynomial $x^{2}+1$
is conjugate to $A$ within ${\rm GL}(2,3)$. Since $B$ also has characteristic (and minimum) polynomial $x^{2}+1,$ it is true that $A$ and $B$ are conjugate within ${\rm GL}(2,3).$
A: You are looking to realize the outer automorphism $i\mapsto j\mapsto k\mapsto i$ of the quaternion group as an inner automorphism inside $GL_2(\Bbb{F}_3)$.
Actually everything happens already inside $SL_2(\Bbb{F}_3)$. This has a normal Sylow $2$-subgroup isomorphic to $Q_8=\langle A,B,C\rangle$. That automorphism is of order three, so we are looking for an element of order three. The first thing that comes to mind
$$
X=\left(\begin{array}{rr}1&1\\0&1\end{array}\right)
$$
happens to work. Namely,
it is easy to verify that
$$
XBX^{-1}=C,\qquad XCX^{-1}=A,\qquad\text{and}\qquad XAX^{-1}=B.
$$
This gives us a realization of $SL_2(\Bbb{F}_3)$ as a semidirect product
$$
SL_2(\Bbb{F}_3)\cong Q_8\rtimes C_3.
$$
A: Let's suppose they are conjugate by 
$$M = \left(\begin{array}{rr}
x & y\\
z & w \end{array} \right).$$
Then, we have
$$MB = AM \Rightarrow \left(\begin{array}{rr}
x + y & x - y\\
z+w & z-w \end{array} \right) = \left(\begin{array}{rr}
-z & -w\\
x & y \end{array} \right).$$
So, $z = -x- y, w = y - x$ and putting these into the last two equations gives
$$x = x, y = y$$
and hence we can choose arbitrary values of $x$ and $y$. So define $M$ as
$$M = \left(\begin{array}{rr}
1 & 0\\
-1 & -1 \end{array} \right).$$
Just for confirmation, 
$$MB = \left(\begin{array}{rr}
1 & 1\\
1 & 0 \end{array} \right)$$
and
$$AM =\left(\begin{array}{rr}
1 & 1\\
1 & 0 \end{array} \right).$$
