I have this proposition and the demonstration, but there is some parts I don't understand, $E$ is a euclidean space and $\Phi$ is a root system with base $\Delta$.
Proposition: Let $\Phi' \subset E'$ be another root system, with base $\Delta'=\{ \alpha_1',\cdots , \alpha_l' \}$. If $<\alpha_i, \alpha_j>=<\alpha_i',\alpha_j'>$ for $1\leq i,j \leq l$, then the bijection $\alpha_i\mapsto \alpha_i'$ extends (uniquely) to an isomorphism $\phi: E \rightarrow E'$ mapping $\Phi$ onto $\Phi'$ and satisfying $<\phi(\alpha),\phi(\beta)>=<\alpha,\beta>$ for all $\alpha,\beta \in \Phi$. Therefore, the Cartan matrix of $\Phi$ determines $\Phi$ up to isomorphism.
Demonstration: Since $\Delta$ (resp. $\Delta'$) is a basis of $E$ (resp. $E'$), there is a unique vector space isomorphism $\phi: E \rightarrow E'$ sending $\alpha_i$ to $\alpha_i'$ with $1\leq i\leq l$. If $\alpha, \beta \in \Delta$, the hypothesis insures that $$\sigma_{\phi(\alpha)}(\phi(\beta))=\sigma_{\alpha'}(\beta')=\beta'-<\beta',\alpha'>\alpha'=\phi(\beta)-<\beta,\alpha>\phi(\alpha)=\phi(\beta-<\beta,\alpha>\alpha)=\phi(\sigma_\alpha(\beta)).$$ In other words, the following diagram commutes for each $\alpha \in \Delta$:
The respective Weyl groups $W, W'$ are generated by simple reflections, so it follows that the map $\sigma \mapsto \phi\circ \sigma\circ \phi^{-1}$ is an isomorphism of $W$ onto $W'$, sending $\sigma_\alpha$ to $\sigma_{\phi(\alpha)}$ where $\alpha \in \Delta$. But each $\beta \in \Phi$ is conjugate under $W$ to a simple root, say $\beta=\sigma(\alpha)$ where $\alpha \in \Delta$. This in turn forces $\phi(\beta)=(\phi\circ \sigma\circ \phi^{-1})(\phi(\alpha)) \in \Phi'$. It follows that $\phi$ maps $\Phi$ to $\Phi'$; moreover, the formula for a reflection shows that $\phi$ preserves all Cartan integers.
The parts I don't understand:
Why "This in turn forces $\phi(\beta)=(\phi\circ \sigma\circ \phi^{-1})(\phi(\alpha)) \in \Phi'$"? I don't get why $(\phi\circ \sigma\circ \phi^{-1})(\phi(\alpha)) \in \Phi'$.
Why "the formula for a reflection shows that $\phi$ preserves all Cartan integers", this is to proof that $<\phi(\alpha),\phi(\beta)>=<\alpha,\beta>$ for all $\alpha,\beta \in \Phi$, don't know how to prove this.
Because $\alpha' := \phi(\alpha) \in \Delta'$ (by assumption $\alpha \in \Delta$), and $w' := (\phi\circ \sigma\circ \phi^{-1}) \in W'$ (shown two sentences before). And of course $w' (\alpha') \in \Phi'$ for each $w' \in W'$ and $\alpha' \in \Delta'$.
"The formula for a reflection" presumably is
$$\sigma_\alpha(\beta) = \beta -\langle \beta, \alpha \rangle \alpha$$
which via linearity of $\phi$ gives
$$\phi(\sigma_\alpha(\beta)) = \phi(\beta) -\langle \beta, \alpha \rangle \phi(\alpha) \qquad(*).$$
We can insert a $\phi^{-1} \circ \phi$ on the left hand side to write $\phi(\sigma_\alpha(\beta)) = \phi(\sigma_\alpha(\phi^{-1}(\phi(\beta)))) = (\phi \circ \sigma_\alpha \circ \phi^{-1})(\phi(\beta))$ which was shown [see Edit below] to be the same as $\sigma_{\phi(\alpha)}(\phi(\beta))$. So we have
$$\sigma_{\phi(\alpha)}(\phi(\beta)) = \phi(\beta) -\langle \beta, \alpha \rangle \phi(\alpha) \qquad (**).$$
But on the other hand, the formula for a reflection on $\Phi'$ says
$$\sigma_{\phi(\alpha)}(\phi(\beta)) = \phi(\beta) -\langle \phi(\beta), \phi(\alpha) \rangle \phi(\alpha) \qquad (***).$$
Compare $(**)$ and $(***)$.
Edit: As correctly pointed out in a comment, the relation $$\phi \circ \sigma_\alpha \circ \phi^{-1} = \sigma_{\phi(\alpha)} \qquad (+)$$ so far has only been shown for $\alpha \in \Delta$, but here we need it for an arbitrary $\alpha \in \Phi$. [The proof actually showed $(\phi \circ \sigma_\alpha \circ \phi^{-1})\color{red}{(\beta)} = \sigma_{\phi(\alpha)} \color{red}{(\beta)}$ for both $\alpha, \beta \in \Delta$, but linearity of both sides immediately implies that this stays true if we replace $\beta$ by any element of $E$ (as is implicitly expressed in the commutative diagram), in particular an arbitrary $\beta \in \Phi$. So we really only need to generalize $(+)$ to arbitrary $\alpha \in \Phi$, which we will do now.]
So let $\alpha \in \Phi$. Again, there is $\alpha_i \in \Delta$ and $w \in W$ such that $w(\alpha_i) = \alpha$. Now it is well known (or easy to see) that $\sigma_\alpha = w \circ \sigma_{\alpha_i} \circ w^{-1}$ and hence
$$\phi \circ \sigma_\alpha \circ \phi^{-1} = \phi \circ w \circ \sigma_{\alpha_i} \circ w^{-1} \circ \phi^{-1} \\= \underbrace{\phi \circ w \circ \phi^{-1}}_{=: w'} \circ \phi \circ \sigma_{\alpha_i} \circ \phi^{-1} \circ \phi\circ w^{-1} \circ \phi^{-1} \\= w' \circ \sigma_{\phi(\alpha_i)} \circ w'^{-1} = \sigma_{w'(\phi(\alpha_i))}\\=\sigma_{\phi(w(\alpha_i))} = \sigma_{\phi(\alpha)}.$$