I've already showed that the map
$$\overbrace{\Bigg\{\begin{pmatrix}a &b\\-b & a\end{pmatrix}:a,b\in\mathbb Z_7\Bigg\}}^{:=G\le \operatorname{GL}_2(\mathbb Z_7)}\times\overbrace{\Bigg\{\begin{pmatrix}x\\y\end{pmatrix}:x,y\in\mathbb Z_7\Bigg\}}^{:=V}\to\Bigg\{\begin{pmatrix}x\\y\end{pmatrix}:x,y\in\mathbb Z_7\Bigg\}\text{ such that}$$
$$\begin{pmatrix}a &b\\-b & a\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}ax+by\\-bx+ay\end{pmatrix} $$
is an action of $G$ on $V$. The order of $G$ is $49$ (so $7^2$ and I could use the relation $|Fix_G(V)|=|V|\text{ (mod 7)}$), which is also the order of the group $V$.
I think that in order to compute the number of orbits of the action I should look at the stabilizer of a certain element in $V$.
For example amatrix $g\in G$ stabilizes an element $\begin{pmatrix}x\\y\end{pmatrix}\iff$
$$g\cdot\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}x\\y\end{pmatrix}\implies\begin{cases} ax+by=x\text{ (mod 7)}\\-bx+ay=y\text{ (mod 7)}\end{cases}(?)$$
but it's not so easy to solve this system. Could you help me please?
Thank you for your attention.
Best Answer
In these kind of cases, it might be helpful to consider a particular element (especially if we take an element with a $0$ entry, it is even easier) and find its stabilizer. For instance, here take $\begin{pmatrix}1\\0\end{pmatrix} \in V$. Then, we have
$$\ \ \ \ a \equiv 1\ (\text{mod } 7) \implies a = 1\\ -b \equiv 0\ (\text{mod } 7) \implies b = 0$$
So, stabilizer of $\begin{pmatrix}1\\0\end{pmatrix}$ consists of a single element, which is the identity of $G$. Thus, by Orbit Stabilizer Theorem, orbit of $\begin{pmatrix}1\\0\end{pmatrix}$ consists of $49 = |V|$ elements, which means there is only $1$ orbit.
Of course, you might think that we were lucky here and we got a stabilizer with $1$ element so obtained the result immediately. If we try to do it by using some Number Theory, one can multiply the first congruence by $y$ and the second one by $x$ to obtain $$\ \ \ \ axy+by^2 \equiv xy\ (\text{mod } 7)\\ \ \ \ \ axy-bx^2 \equiv xy\ (\text{mod } 7)$$ which gives us $$b(x^2+y^2) \equiv 0\ (\text{mod } 7)$$
and we should use Sum of two squares Theorem to obtain $b = 0$ (note that $x^2+y^2 = 7$ has no integer solutions since $7 \equiv 3\ (\text{mod }4)$). Then again, the result you mentioned in comments follows.