The fibres of (trace, determinant) on 2×2 matrices

The fibre. Fix $s=\operatorname{tr}$ and $\delta=\det$, and put $\Delta=s^2-4\delta$. In the coordinates $(u,v,w)\in\mathbb{R}^3$ the fibre is the quadric
$$Q_\Delta=\{u^2+v^2-w^2=\Delta\}.$$

Case $\Delta>0$ (two distinct real eigenvalues). $Q_\Delta$ is a one-sheeted hyperboloid, connected. All such matrices are diagonalisable with the same eigenvalues $\frac{s\pm\sqrt\Delta}2$, hence all conjugate: the fibre is a single conjugacy class.

Case $\Delta<0$ (complex conjugate eigenvalues). $w^2=u^2+v^2+|\Delta|$: a two-sheeted hyperboloid, with two components $w>0$ and $w<0$. Over $\mathbb{R}$, all matrices with the same non-real eigenvalues are similar (rational canonical form), so the fibre is one $GL_2(\mathbb{R})$-orbit. It is disconnected: conjugation by $\operatorname{diag}(1,-1)$ (determinant $-1$) sends $(b,c)\mapsto(-b,-c)$, hence $w\mapsto-w$, swapping the two sheets. Under the connected subgroup $GL_2^+(\mathbb{R})$ each sheet is an orbit.

Case $\Delta=0$ (double eigenvalue $\frac s2$). $u^2+v^2=w^2$: a double cone. The apex $u=v=w=0$ is the scalar matrix $\frac s2I$, a one-point conjugacy class. The rest, the cone minus the apex (two nappes, $w>0$ and $w<0$), consists of the matrices $\frac s2I+N$ with $N\ne0$ nilpotent. All nonzero nilpotent $2\times2$ matrices are similar to $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, so this is a single $GL_2(\mathbb{R})$-orbit (again with two components, swapped by $\operatorname{diag}(1,-1)$).

Summary. Each fibre with $\Delta\ne0$ is one conjugacy class, and the fibre with $\Delta=0$ consists of two: the scalar matrix and the non-diagonalisable matrices.

Check. $\Delta=0$, $s=2$, $\delta=1$: $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has $u=0$, $v=1$, $w=1$: on the cone. The identity $I$ has $u=v=w=0$: the apex.

Remark. The orbit space $M_2(\mathbb{R})/GL_2(\mathbb{R})$ is *not* the plane $\mathbb{R}^2$ of $(s,\delta)$: the quotient is non-Hausdorff along the parabola $\Delta=0$ (the apex and the cone minus apex cannot be separated), while the map $(\operatorname{tr},\det)$ identifies them.