There are many ways of making the additive group $\Bbb{Z}^n$ (where $n \in \Bbb{N}$) into a totally ordered group. E.g., if $n = 2$, and $\rho \in \Bbb{R} \setminus \Bbb{Q}$, you can get a totally ordered abelian group of rank $2$, $G(\rho)$ say, by identifying $\Bbb{Z}^2$ with a subgroup of $\Bbb{R}$ via $(m, n) \mapsto m + n\rho.$ Then $G(\rho) \cong G(\sigma)$ iff $\sigma = (c + d\rho)/(a + b\rho)$, where $a, b, c, d \in \Bbb{Z}$ with $ad - bc = \pm 1$. (See Lord Shark's answer below). Also the $G(\rho)$ and the lexicographic order give all the possibilities for $n = 2$. (Please correct me if am wrong about this.)
Is there a classification of totally ordered finitely generated abelian groups of rank $n$ for $n > 2$? (Or for $n = 2$, if my ideas above are wrong?) I'd also be very interested in pointers to any partial results if a general classification is not known.
There are other possibilities for isomorphism beyond $\rho/\sigma$ being rational. We can rewrite the definition of $G(\rho)$ by saying that it is a free group with generators $e_1$ and $e_2$ with $me_1+ne_2>0$ iff $m+n\rho>0$. We can choose new generators $f_1$ and $f_2$ by setting $f_1=ae_1+be_2$ and $f_2=ce_1+de_2$ where $a,\ldots,d$ are integers with $ad-bc=\pm1$. Then $mf_1+nf_2>0$ iff $ma+nc+(mb+nd)\rho>0$, that is $m(a+b\rho)+n(c+d\rho)>0$. In the case where $a+b\rho>0$ this is then $G(\sigma)$ with $\sigma=(c+d\rho)/(a+b\rho)$.