In my opinion, it's probably not so interesting to consider all extensions. 
If you consider the Lie subalgebra $\mathfrak{n}_n\subset \mathfrak{gl}_n$ of all strictly upper triangular matrices, then $\mathfrak{n}_n$ is an extension of $\mathbf{C}$ because it has a quotient Lie algebra of dimension 1. It follows that even for nilpotent extensions the nilpotency can grow arbitrarily large.
Of course extensions can be even more chaotic than that. 


**Edit**
Example of a non-central perfect group extension: Consider block matrices over $\mathrm{C}$
$$
 H = \begin{pmatrix}\mathrm{SL_2} & * \\ 0 & \mathrm{SL_2}\end{pmatrix},\;
 K = \begin{pmatrix}I_2 & * \\ 0 & I_2\end{pmatrix},\; G = H/K \cong \mathrm{SL}_2\times \mathrm{SL}_2
$$
The group $H$ is perfect because its Lie algebra is perfect: The diagonal part $\mathfrak{sl}_2\times \mathfrak{sl}_2$ lies in the commutator because $\mathfrak{sl}_2$ is perfect. On the other hand, we have
$$
X = \begin{pmatrix}0 & 0 & b & a \\ 0 & 0 & d & c \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}, Y = \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix},\; [X, Y] = \begin{pmatrix}0 & 0 & a & b \\ 0 & 0 & c & d \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{pmatrix}
$$
so the off-diagonal part is also in the commutator.

Consider the conjugation by the matrix $\mathrm{diag}(1, 1, -1, -1)$. It acts non-trivially on $H$ but trivially on $G$.
It means that non-central extensions can in general have non-trivial automorphisms even if the extension is a perfect group.