The Cartan matrix determines a based root system

Every root of $\mathrm{\Phi }$ is Weyl-conjugate to a simple root. Indeed, let $\gamma ={\sum }_i{n}_i{\alpha }_i$ be positive and not simple. Since $0<(\gamma ,\gamma )={\sum }_i{n}_i(\gamma ,{\alpha }_i)$, some $i$ satisfies $(\gamma ,{\alpha }_i)>0$. Put $m=2(\gamma ,{\alpha }_i)\mathrm{/}({\alpha }_i,{\alpha }_i)>0$. Crystallographic integrality gives $m\in \mathbb{Z}$, and reflection invariance gives $s_{{\alpha }_i}(\gamma )=\gamma -m{\alpha }_i\in \mathrm{\Phi }$. Its height is $\mathrm{ht}(\gamma )-m<\mathrm{ht}(\gamma )$. It cannot be negative: if it were, then all its simple-root coefficients would be nonpositive by [L1], whereas its coefficient at every $j\mathrm{\ne }i$ is $n_j\ge 0$; hence all $n_j$ for $j\mathrm{\ne }i$ would vanish, making $\gamma$ a positive scalar multiple of ${\alpha }_i$, and reducedness would force $\gamma ={\alpha }_i$, contrary to assumption. Thus successive simple reflections strictly lower positive height until a simple root is reached. Inverting those reflections proves the claim.

With the row index first and column index second, the matrix of $s_{{\alpha }_i}$ in the basis $\mathrm{\Delta }$ has entries $(s_{{\alpha }_i}{)}_{jk}={\delta }_{jk}-a_{ik}{\delta }_{ji}$ by [L2]; it is therefore determined by $A$, and the corresponding matrix for $s_{{\alpha }_i^{\mathrm{\prime }}}$ in the basis ${\mathrm{\Delta }}^{\mathrm{\prime }}$ is the same. Hence $\phi s_{{\alpha }_i}{\phi }^{-1}=s_{{\alpha }_i^{\mathrm{\prime }}}$ for all $i$, because both sides are linear and agree on the basis ${\mathrm{\Delta }}^{\mathrm{\prime }}$: $\phi s_{{\alpha }_i}{\phi }^{-1}({\alpha }_j^{\mathrm{\prime }})=\phi ({\alpha }_j-a_{ij}{\alpha }_i)={\alpha }_j^{\mathrm{\prime }}-a_{ij}{\alpha }_i^{\mathrm{\prime }}=s_{{\alpha }_i^{\mathrm{\prime }}}({\alpha }_j^{\mathrm{\prime }})$.

Consequently $\phi$ carries the Weyl orbit of $\mathrm{\Delta }$ onto the Weyl orbit of ${\mathrm{\Delta }}^{\mathrm{\prime }}$, and by step 1.1 (applied to $\mathrm{\Phi }$ and to ${\mathrm{\Phi }}^{\mathrm{\prime }}$) it carries $\mathrm{\Phi }$ onto ${\mathrm{\Phi }}^{\mathrm{\prime }}$; in particular $\phi$ is a linear isomorphism, since it maps the basis $\mathrm{\Delta }$ onto the basis ${\mathrm{\Delta }}^{\mathrm{\prime }}$.

It remains to check that $\phi$ preserves Cartan integers. By [L3] the Gram matrices $G$ and $G^{\mathrm{\prime }}$ satisfy $a_{ij}=2G_{ij}\mathrm{/}{G}_{ii}=2G_{ij}^{\mathrm{\prime }}\mathrm{/}{G}_{ii}^{\mathrm{\prime }}$ and are positive definite; for indices $i,j$ joined by an edge one has $a_{ij},a_{ji}<0$ and $a_{ij}{G}_{ii}=2G_{ij}=2G_{ji}=a_{ji}{G}_{jj}$, so $G_{ii}^{\mathrm{\prime }}\mathrm{/}{G}_{ii}=G_{jj}^{\mathrm{\prime }}\mathrm{/}{G}_{jj}$; by connectivity along edges the ratios $G_{ii}^{\mathrm{\prime }}\mathrm{/}{G}_{ii}$ are constant on each connected component of the graph on $\mathrm{\Delta }$ with edges $a_{ij}\mathrm{\ne }0$. Define a second inner product on $E$ by $(x,y{)}_1=(\phi x,\phi y{)}_{E^{\mathrm{\prime }}}$. Its Gram matrix in the basis $\mathrm{\Delta }$ is $G^{\mathrm{\prime }}$, which differs from $G$ by a positive scalar on each connected component; therefore for roots $\gamma ,\delta$ of $\mathrm{\Phi }$ the Cartan integers computed with $(\cdot ,\cdot )$ and with $(\cdot ,\cdot {)}_1$ agree, because scaling an inner product on a component by $\lambda >0$ multiplies both $(\gamma ,\delta )$ and $(\delta ,\delta )$ by $\lambda$ when $\gamma ,\delta$ lie in that component and gives $0$ otherwise. Since $\phi (\mathrm{\Phi })={\mathrm{\Phi }}^{\mathrm{\prime }}$ by step 2.1, $\phi$ is an isomorphism of root systems.