Where is the mistake in the following argument? I am arguing that the affine group on $\mathbb{R}^n$ is unimodular, which it is not!.
Let $$\operatorname{Aff}_n(\mathbb{R}) := \left\{\begin{pmatrix} A & v\\ 0 & 1 \end{pmatrix}: \, A \in \operatorname{GL}_n(\mathbb{R}), v \in \mathbb{R}^n\right\}$$ be the group of affine transformations on $\mathbb{R}^n$. Denote $h := \begin{pmatrix} A_h & v_h \\ 0 & 1 \end{pmatrix}$.
Then I define $f: \operatorname{Aff}_n(\mathbb{R}) \rightarrow \mathbb{R}_{\geq 0}$ via $h \mapsto \vert \det A_h \vert^{-(n+1)}$ and claim that the measure $\mu$ having density $f$ w.r.t. the $n+1$-dimensional Lebesgue measure $\lambda^{n+1}$ is left and right translation invariant.
Let $g \in \operatorname{Aff}_n(\mathbb{R})$ and $B$ a Borel set of $\operatorname{Aff}_n(\mathbb{R})$ be arbitrary. Then by definition
$$ \mu(gB) = \int_{gB} f(h) d\lambda^{n+1}(h)$$
Now note that $gB = g_l (B)$ where $g_l: G \rightarrow G$ via $g_l(h) = gh$. This function is smooth, so one can apply the Jacobi transformation formula to obtain
$$\int_{B} \vert \det D(g_l) \vert f(g_l(h)) d\lambda^{n+1}(h)$$
Now note that $D(g_l) = I_n \otimes g$ and $f(g_l(h)) = f(gh)$ which gives
$$\int_{B} \vert \det A_g \vert^{n+1} f(gh) d\lambda^{n+1}(h)$$
Now we use that
$$\vert \det A_{gh} \vert^{-(n+1)} = \vert \det A_h \vert^{-(n+1)} \vert \det A_g \vert^{-(n+1)}$$
to finally get
$$\int_{B} \vert \det A_h \vert^{-(n+1)} d\lambda^{n+1}(h) = \int_{B} f(h) d\lambda^{n+1}(h) = \mu(B)$$
Now we make the same argument for $Bg$ i.e. $g_r(B)$ where $g_r(h) = hg$. Then the only difference is that $D(g_r) = g^T \otimes I_n$, but the determinant of this matrix is also $\vert \det A_g \vert^{(n+1)}$. Also for the second term in the integrand we have
\begin{align} f(g_r(h)) = f(hg) &= \vert \det A_{hg} \vert^{-(n+1)} \\ &= \vert \det (A_{h} A_g) \vert^{-(n+1)} \\ &= \vert \det A_{h} \det A_g \vert^{-(n+1)} \\ &= \vert \det A_{h} \vert^{-(n+1)} \vert \det A_g \vert^{-(n+1)} \\ \end{align}
which is the same as before. Hence also $\mu(Bg) = \mu(B)$ which means that the left and right Haar measure conicide.
So the question is: Where do I go wrong?
Any comments, suggestions or references are very much appreciated too. Thank you in advance.
Note that you need to consider the $(n+1)^2$-dimensional Lebesgue measure here since you embed $\operatorname{Aff}_n(\mathbb{R})$ into $\mathbb{R}^{(n+1) \times (n+1)} \cong \mathbb{R}^{(n+1)^2}$.
Otherwise the calculations are correct, but the constructed left and right translation invariant measure $\mu$ is trivial because $\operatorname{Aff}_n(\mathbb{R})$ is a null set:$$\lambda^{(n+1)^2}(\operatorname{Aff}_n(\mathbb{R}) )\leq \lambda^{(n+1)^2}(\mathbb{R}^{n^2+n} \times \{0\}^n \times \{1\}) = 0.$$ Hence $\mu$ is not a Haar measure and your argument does not contradict the fact that $\operatorname{Aff}_n(\mathbb{R})$ is not unimodular.