$\newcommand{\d}{\mathrm{d}}$ This the Exercise 3, Chapter 11 of the Gerald B. Folland book Real Analysis:
Let $G$ be a locally compact group that is homeomorphic to an open subset $U$ of $\Bbb R^d$ in such a way that, if we identify $G$ with $U$, left translation is an affine map -- that is, $xy=A_x(y)+b_x$ where $A_x$ is a linear transformation of $\Bbb R^d$ and $b_x\in\Bbb R^d$. Then $|\det A_x|^{-1}\d x$ is a left Haar measure on $G$, where $\d x$ denotes Lebesgue measure on $\Bbb R^d$.
Let me know if I understand this.
What the problem gives us is $G$ a locally compact group, an open set $U\subseteq\Bbb R^d$, and a bijection $\varphi:G\to U$ such that $\varphi$ and $\varphi^{-1}$ are both continuous and satisfy that given $u\in G$ there is a linear operator $A_{\varphi(u)}:\Bbb R^d\to \Bbb R^d$ and $b_{\varphi(u)}\in\Bbb R^d$ so that $\varphi(uv)=A_{\varphi(u)}(\varphi(v))+b_{\varphi(u)}.$
Is this right
If it is, define $f:\Bbb R^d\to\Bbb [0,\infty[$ given by $f(x)=|\det A_x|^{-1}\quad\text{i.e.}\quad f(x)=|\det A_{\varphi(\varphi^{-1}(x))}|^{-1}.$
Is the problem asking if the measure $\mu$ in $G$ given by $\mu(E)=\int_{\varphi(E)}f(x)\d x$ is a left Haar measure?
This reminds me the formula $\int_E f(y)\d y=|\det T|\int_{T^{-1} E} f(Tx)\d x,$ but I don't know what can I do.
Any advice in how to interpret the problem or on how to proceed is very appreciated.