How can one assume that an isomorphism of root spaces $\Phi\to\Phi'$ comes from an isometry? By definition, if $\Phi$ and $\Phi'$ are root systems of the Euclidean spaces $E$ and $E'$, respectively, then an isomorphism $\Phi\to\Phi'$ is one that is induced by an isomorphism $E\to E'$ which preserves the inner product on pairs of roots, but is not necessarily an isometry. 
I read that since the axioms of a root system are unchanged by scaling the inner products be a positive real number, we can assume that the isomorphism is induced by an isometry. 
How is this possible? It seems like this suggests that given an isomorphism $\psi\colon E\to E'$, it's possible to scale the inner products on $E$ and $E'$ so that $s(\psi(x),\psi(y))=r(x,y)$ for all $x,y\in E$ and some fixed $s,r>0$. But this doesn't seem like it'd be true at all.
 A: I only know the case where $\Phi,\Phi'$ are assumed to be irreducible root systems.
Recall that for irreducible systems, roots of the same length are conjugate under the Weyl group $\mathcal{W}$.
(These can be found in Humphreys' Introduction to Lie Algebras and Representation Theory, Section 10.4)
Since and isomorphism of roots system comes from an isomorphism of vector space preserving the $\langle \ , \ \rangle$ operation, so in order to make the map $\pi:E\to E'$ an isometry, we only need to show that
$$(\alpha,\alpha)_E=(\pi\alpha,\pi\alpha)_{E'}$$
for each $\alpha\in E$. Since for each $\sigma\in\mathcal{W}$ that
$$(\sigma\alpha,\sigma\alpha)_E=(\alpha,\alpha)_E$$
(similarly for $\mathcal{W}',E'$), we see that for $\alpha,\beta\in\Phi$, we have
$$(\alpha,\alpha)_E=(\beta,\beta)_E\quad\Rightarrow\quad(\pi\alpha,\pi\alpha)_{E'}=(\pi\beta,\pi\beta)_{E'}$$
For roots of distinct lengths, notice
$$\frac{(\pi\alpha,\pi\alpha)_{E'}}{(\pi\beta,\pi\beta)_{E'}}=\frac{\langle \pi\alpha,\pi\beta \rangle _{E'}}{\langle \pi\beta,\pi\alpha \rangle _{E'}}=\frac{\langle \alpha,\beta\rangle_{E}}{\langle \beta,\alpha\rangle_{E}}=\frac{(\alpha,\alpha)_E}{(\beta,\beta)_E}$$
(Warning : for this equation to make sense, we need to assume that $(\alpha,\beta)\neq 0$)
These together shows the assertion.
