Lie algebra of the automorphism group

Choose a basis of $\mathfrak{g}$. The equations $A[x,y]=[Ax,Ay]$ for basis pairs are finitely many polynomial equations in the matrix entries of $A$. Their common zero set inside $\mathrm{GL}(\mathfrak{g})$ is exactly $\mathrm{Aut}(\mathfrak{g})$ and is closed. By [L1] it is an embedded Lie subgroup; for a complex algebra, apply [L1] first to the underlying real group.

A tangent vector at the identity is represented by $A(t)=I+tD+o(t)$. Differentiating the bracket equation gives $D[x,y]=[Dx,y]+[x,Dy]$, so the tangent algebra is contained in $\mathrm{Der}(\mathfrak{g})$. Conversely, for a derivation $D$, the linear vector field $A\mapsto DA$ is tangent to the defining equations, or equivalently its local flow $\exp (tD)$ preserves the bracket by differentiating $\exp (tD)[x,y]-[\exp (tD)x,\exp (tD)y]$; hence every derivation is tangent. When $\mathfrak{g}$ is complex, the ambient group consists of complex-linear maps and the resulting space of complex-linear derivations is stable under multiplication by $i$; exponential charts therefore give the embedded subgroup its corresponding complex Lie-subgroup structure.

Step 1.2 identifies the Lie algebra with $\mathrm{Der}(\mathfrak{g})$, and [L2] identifies that with $\mathrm{ad}(\mathfrak{g})$. If $\mathfrak{g}=0$, the automorphism group is the one-point group and all tangent algebras are zero. Countable choice is used only through [L1], not in the polynomial or differentiation steps.