In the Faddeev-Popov procedure one defines the Faddeev-Popov determinant through the formula $$\int {\mathcal{D}\alpha \ } \delta\big[G(A^\alpha)\big]\Delta[A]=1,\tag{1}$$
where $G(A^\alpha)$ is the gauge-fixing condition and $A^\alpha$ is the gauge field $A$ transformed by a finite gauge transformation obtained exponentiating $\alpha$:
$$(A^\alpha)^a_\mu T^a=e^{i\alpha^a T^a}A^b_\mu T^b e^{-i \alpha^c T^c}+\dfrac{i}{g}e^{i\alpha^a T^a}\partial_\mu e^{-i\alpha^b T^b}.\tag{2}$$
Question: Is this $\cal D \alpha$ measure related to the Haar measure on the gauge group ${\cal G}$? In other words, is (1) exactly the same as $${\int \mathcal{D}g \ }\delta\big[G(A^g)\big]\Delta[A]=1\tag{3}$$
where now $$A^g=g Ag^{-1}+\dfrac{i}{e} g \partial_\mu g^{-1}\tag{4}$$
where in (4) the measure ${\cal D}g$ is the Haar measure? What is the precise relation between $\cal D\alpha$ and $\cal D g$?
I mean, heuristically (1) and (3) are doing the exact same thing, since by varying all $\alpha$ we vary all $g$ and since $A^\alpha = A^{g(\alpha)}$ where $g(\alpha)=e^{i\alpha^a T^a}$, but it is not clear to me if the measure ${\cal D}\alpha$ is related (and how it is related) to the Haar measure ${\cal D}g$.
My intuition here is that this is analogous to what we do with the Lebesgue measure $d^nx$ in $\mathbb{R}^n$ when we express it in various coordinate systems. We can express it in cartesian, spherical or cylindrical coordinates but it is always the same measure. My intuition is that here we are using the functions $\alpha$ as the "coordinates" and in (1) we are just writing down the Haar measure in these coordinates. Is that correct?