We have the general linear group $\mathrm{GL}_{2}(\mathbb{C})$ that acts on the set $\mathrm{M}_2(\mathbb{C})$ of $(2 \times 2)$-matrices by conjugation.
I want to classify the orbits of this action.
What I know: $$ \mathrm{Orb}_{G}(A) = \{ BAB^{-1} : B \in \mathrm{GL}_2(\mathbb{C})\} \,. $$
On
For $A\in M_2(\Bbb{C})$ the minimal polynomial determinates the conjugacy class $\{BAB^{-1},B\in GL_2(\Bbb{C})\}$.
If the minimal polynomial is $X-a$ then the matrix is $A=aI$.
If the minimal polynomial is $X^2+aX+b$ then take $v\in \Bbb{C}^2$ not an eigenvector so that $\Bbb{C}^2 = \Bbb{C}v\oplus \Bbb{C} Av$ and the matrix acts as $A(cv+dAv) = (c-ad)Av-bd v$.
As already mentioned in the comments this can be solved via the Jordan normal form. But in the special case of $2 \times 2$ matrices this can also be done by hand: let $A \in \mathrm{M}_2(\mathbb{C})$ and let $\lambda$ and $\mu$ be the eigenvalues of $A$.
If $\lambda \neq \mu$, then all eigenvalues of $A$ are distinct, so $A$ is diagonalizable. Thus $A$ is conjugated to a matrix of the form $$ \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}. $$
If $\lambda = \mu$ and $A$ is still diagonalizable, then $A$ is conjugated to the matrix $$ \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}. $$
If $\lambda = \mu$ but $A$ is not diagonalizable, then $A$ can at least be triangularized (because $\mathbb{C}$ is algebraically closed), so $A$ is conjugated to a matrix $$ \begin{pmatrix} \lambda & a \\ 0 & \lambda \end{pmatrix} $$ with $a \neq 0$. By further conjugating with the diagonal matrix $\begin{pmatrix} a^{-1} & 0 \\ 0 & 1 \end{pmatrix}$ we find that $A$ is already conjugated to the matrix $$ \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}. $$
We find overall that the following three types of matrices represent all conjugation classes: $$ \text{ $\begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}$ with $\lambda \neq \mu$} \,, \quad \text{ $\begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix}$ with $\lambda \in \mathbb{C}$} \,, \quad \text{ $\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$ with $\lambda \in \mathbb{C}$} \,. $$ We need to examine if there is some redundancy between these representatives.
No two matrices of different types are conjugated to one another, because they are characterized by properties independent under conjugation (diagonalizable with distinct eigenvalues, diagonalizable with same eigenvalues, not diagonalizable). So we only need to examine if two matrices of the same type can be conjugated to one another.
For matrices of the second and third type this is not the case because the eigenvalues are constant on the conjugation classes. For the first type we have precisely two representatives for each conjugation class, because we can swap the two diagonal entries.
Therefore, we can represent the conjugation classes by the set $$ \left\{ \begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix} \middle|\, \lambda, \mu \in \mathbb{C}, \lambda \neq \mu \middle\}\middle/{\sim} \right. \quad\cup\quad \left\{ \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} \,\middle|\, \lambda \in \mathbb{C} \right\} \quad\cup\quad \left\{ \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix} \,\middle|\, \lambda \in \mathbb{C} \right\} \,, $$ where $$ \begin{pmatrix} \lambda_1 & 0 \\ 0 & \mu_1 \end{pmatrix} \sim \begin{pmatrix} \lambda_2 & 0 \\ 0 & \mu_2 \end{pmatrix} $$ if and only if $\lambda_1 = \lambda_2$, $\mu_1 = \mu_2$ or $\lambda_1 = \mu_2$, $\mu_1 = \lambda_2$.