Let $G$ be a metric group and let $\sA ut(G)$ denote the automorphism group of $G$. If $\sA$ and $\sB$ are groups of $G$-valued maps defined on the sets $X$ and $Y$, respectively, we say that $\sA$ and $\sB$ are \emph{equivalent} if there is a group isomorphism $H\colon\sA\to\sB$ such that there is a bijective map $h\colon Y\to X$ and a map $w\colon Y\to \sA ut (G)$ satisfying $Hf(y)=w[y](f(h(y)))$ for all $y\in Y$ and $f\in \sA$. In this case, we say that $H$ is represented as a \emph{weighted composition operator}. A group isomorphism $H$ defined between $\sA$ and $\sB$ is called \emph{separating} when for each pair of maps $f,g\in \sA$ satisfying that $f^{-1}(e_G)\cup g^{-1}(e_G)=X$, it holds that $(Hf)^{-1}(e_G)\cup (Hg)^{-1}(e_G)=Y$. Our main result establishes that under some mild conditions, every separating group isomorphism can be represented as a weighted composition operator. As a consequence we establish the equivalence of two function groups if there is a biseparating isomorphism defined between them.