Suppose $\psi: GL_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ defined by sending $$ g\mapsto gAg^{-1}. $$
Then why is it that $d\psi:T_eGL_2(\mathbb{C})=M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ is defined to be $$ C\mapsto [C,A]? $$
It is sort of related to this link, but I am not sure if the same strategy as in that link will work here.
The same strategy works (with $2$ replaced by $n$). The saving grace here is that $\text{GL}_n(\mathbb{C})$ naturally embeds into $\mathcal{M}_n(\mathbb{C})$ and its tangent space at any point can be canonically identified with $\mathcal{M}_n(\mathbb{C})$. So write $g = 1 + \epsilon C$ and compute the $\epsilon$ term of $gAg^{-1}$...