It is known that homeomorphisms in general do not preserve null sets.
However, I am wondering about the case where $G$ is a locally compact group with haar measure $\mu$ and the homeomorphism in question is the inverse map.
Is the following true for a borel subset $A$? $$ 0<\mu (A)<\infty \Longleftrightarrow 0<\mu (A^{-1})<\infty $$
It is true that $\mu(A)>0$ iff $\mu(A^{-1})>0$. Here is a sketch of a proof. Let $\nu(A)=\mu(A^{-1})$. Since $\mu$ is a (left) Haar measure, $\nu$ is a right Haar measure. On the other hand, for any $x\in G$, note that $A\mapsto \mu(Ax)$ is another left Haar measure, and so by uniqueness of Haar measure there is some constant $\Delta(x)>0$ such that $\mu(Ax)=\Delta(x)\mu(A)$ for all $A$. You can now check that $\Delta:G\to (0,\infty)$ is a continuous homomorphism, and that the measure $\Delta^{-1} \mu$ (i.e., $A\mapsto \int_A \Delta^{-1}\,d\mu$) is a right Haar measure. But that means $\nu$ is a constant multiple of $\Delta^{-1}\mu$ (in fact they can be shown to be equal). In particular, since $\Delta^{-1}$ is always positive, this implies $\mu(A)>0$ iff $\nu(A)>0$.
It is not true in general that $\mu(A)<\infty$ iff $\nu(A)<\infty$, though. The function $\Delta$ may get arbitrarily small (and not just on sets whose measure is going to $0$ faster), and so you can get a set $A$ such that $\mu(A)$ is finite but $\Delta$ approaches zero fast enough on $A$ that $\nu(A)=\infty$.
For an explicit example, let $G$ be the group of invertible maps $\mathbb{R}^2\to\mathbb{R}^2$ of the form $x\mapsto ax+b$, topologized by identifying $G$ with $\{(a,b)\in\mathbb{R}^2:a\neq 0\}$. Then if $\lambda$ is Lebesgue measure on $\mathbb{R}^2$, it turns out that a left Haar measure on $G$ is $\mu=\frac{1}{a^2}\lambda$ and the corresponding right Haar measure is $\nu=\frac{1}{|a|}\lambda$. So, for instance, if $A=\{(a,b):a>1,0<b<1\}$, then $\mu(A)=\int_1^\infty\frac{1}{a^2}\,da<\infty$ but $\nu(A)=\int_1^\infty\frac{1}{|a|}\,da=\infty$.
This function $\Delta$ is known as the modular function of $G$.