Number of orbits of an action 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.
 A: 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.
