Endomorphisms preserve Haar measure
Solution 1:
The moral is that you have to assume surjectivity of $A$.
For the middle equality: we have $A^{-1}(Ax.E)=x.A^{-1}E$ as sets:
$y\in A^{-1}(Ax.E)$ means $Ay\in Ax.E$ means $A(x^{-1}y)\in E$, while
$y\in x.A^{-1}E$ means $x^{-1}y\in A^{-1}E$.
For the rotation-invariance: we now see $\mu(Ax.E)=\mu(E)$ for all $x\in G$. In other words, $\mu(y.E)=\mu(E)$ for all $y\in \text{im} A\subset G$. By surjectivity we have $\text{im} A=G$.