It is given in [Regular left-orders on groups][1] that the solvable Baumslag-Solitar group $\mathrm{BS}(1,n)=\langle a, b\mid aba^{-1}=b^n\rangle $ is isomorphic to  $\mathbb{Z}[1/n]\rtimes \mathbb{Z}$ for $n\in \mathbb{Z}.$ Can we find an isomorphism $\phi:\mathbb{Z}[1/n]\rtimes \mathbb{Z}\rightarrow \mathrm{BS}(1,n) $ such that if $u<v$ implies $\phi(u)<\phi(v),$ where $u,v\in\mathbb{Z}[1/n]\rtimes \mathbb{Z}$. 

Let $u=(r,m)$ and $v=(r',m')$ so $u<v$ means $(r,m)<(r',m')$ which happens if either $m'>m$ or if $m=m'$ then $r'>r.$ In BS(1,n) every element can be written as can be
written in the form $a^n(a^{-m}b^ka^m)$with $m\geq 0$, $n, k \in \mathbb{Z}$. So $a^{n_1}(a^{-{m_1}}b^{k_1}a^{m_1})>a^{n_2}(a^{-{m_2}}b^{k_2}a^{m_2})$ if either $n_1>n_2$ or if $n_1=n_2$ then $m_1>m_2.$



  [1]: https://arxiv.org/abs/2104.04475