I am trying to solve this problem.
Let $G$ be $GL_2(\mathbb{F}_3)$ and $$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, B = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, C=\begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}, X=\{\{\pm A\}, \{\pm B\}, \{\pm C\}\}$$. Show that conjugation induces an group action by $G$ on $X$.
Since each set in $X$ contains $\pm$, it is enough to check $M \in G$ such that $\det M = 1$.
Write $ M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $$M^{-1}=\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$$
Then, $$MAM^{-1} = \begin{pmatrix} ac+bd & -b^2-a^2 \\ c^2+d^2 & -ac-bd \end{pmatrix}$$ $$MBM^{-1} = \begin{pmatrix} ad+bd-ac+bc & a^2-2ab-b^2 \\ 2cd+d^2-c^2 & -bc-bd+ac-ad \end{pmatrix}$$ $$MCM^{-1} = \begin{pmatrix} -ad+bd-ac-bc & 2ab-b^2+a^2 \\ -2cd-c^2+d^2 & bc-bd+ac+ad \end{pmatrix}$$
So I can check that the diagonal entries are - each other. But I don't know how to check the off diagonal components are equal in the cases of $MBM^{-1}, MCM^{-1}$ and - in the case of $MAM^{-1}$.
Is there any clever method to check this other than calculating all the elements of $G$?