$\newcommand{\mmod}{/\!\!/}\newcommand{\fr}[1]{\mathfrak{#1}}$ If $G$ is a torus, then $G$ acts trivially on $\fr{g}$, so that $\fr{g}\mmod G \cong \fr{g}$. Similarly, the adjoint action of $G$ on $G$ is trivial, so that $G\mmod G \cong G$, which is of course not isomorphic to $\fr{g}$. Suppose instead that $G$ is semisimple and simply-connected (everything over an algebraically closed field $k$ of characteristic zero, to be safe), and let $W$ be the Weyl group. Then $\fr{g}\mmod G \cong \fr{t}\mmod W$ by Chevalley, and $G\mmod G \cong T\mmod W$. Both $\fr{t}\mmod W$ and $T\mmod W$ are both isomorphic to an affine space of dimension $\mathrm{rank}(G)$; for $\fr{t}\mmod W$, this is Chevalley-Shepard-Todd, and for $T\mmod W$, this is Theorem 6.1 of Steinberg's "Regular elements of semisimple algebraic groups". So they are indeed both isomorphic. For example, say $G = \mathrm{SL}_2$; then $\fr{t}\mmod W \cong \mathrm{Spec}(k[x^2])$ and $T\mmod W \cong \mathrm{Spec}(k[y + y^{-1}])$, and the isomorphism sends $x^2\mapsto y + y^{-1}$.
I don't know what happens if $G$ is instead unipotent, but going through the references of the paper you linked, I found https://arxiv.org/pdf/1203.3000.pdf. In the introduction, the author writes that if $N$ is the subgroup of $\mathrm{GL}_n(k)$ of upper triangular matrices with $1$s on the diagonal and $\fr{n}$ is its Lie algebra, then $k[\fr{n}]^N$ is a polynomial algebra on $n-1$ variables. (So $\fr{n}\mmod N$ is an affine space of dimension $n-1$.)