Ben, I also asked myself that same question and the notion that made it clear for me was that of an isomorphism between root systems: it is a linear map that sends all the roots of one system to all the roots of the other system, preserving the Cartan-Killing numbers of the corresponding roots (see Humphreys' book, Section 9.2)

The beauty is that such an isomorphism does not need to be an isometry! So the the infinitely many two-dimensional root systems of the 90$^o$ case are all isomorphic between themselves, in particular all isomorphic to $A_1 \times A_1$.

In terms of the corresponding Lie algebras, you have also infinitely many, all of them isomorphic to ${\frak sl}(2) \times {\frak sl}(2)$.

PS - It can be shown that an isomorphism between two *irreducible* root systems must be conformal: it must scale the metrics by a constant factor (proof below). In particular, every automorphism of an irreducible root system is an isometry.
The irreducibility here is crucial since an $A_1 \times A_1$ root system with distinct root lengths admit an automorphism which is not an isometry.

This proof is in "Álgebras de Lie" by Luiz San Martin, Editora da UNICAMP but I will use Humphreys's notation instead.

**Lemma:** In an irreducible root system $\Phi$, given any pair of roots $\alpha$, $\beta$ there exists a sequence of roots $\alpha = \alpha_1, \, \alpha_2,\, \ldots,\, \alpha_n = \beta$ which are consecutively non-orthogonal, that is, such that $(\alpha_i, \alpha_{i+1}) \neq 0$, for $i=1,2,\ldots,n-1$.

*Proof:* Write $\alpha \sim \beta$ if there is such a sequence of roots connecting $\alpha$ and $\beta$ and $\alpha \not\sim \beta$ on the contrary. Consider $\Phi_\alpha = \{ x \in \Phi:\, \alpha \sim x \} $ and $\Phi^0_\alpha = \{ y \in \Phi:\, \alpha \not\sim y \} $. These subsets are orthogonal since if a $y \Phi^0_\alpha $ was not orthogonal to a $x \in \Phi_\alpha$ then $\alpha \sim x$ and $x \sim y$ would imply $\alpha \sim y$. Since $\alpha \in \Phi_\alpha$ and $\Phi$ is irreducible it follows that $\Phi^0_\alpha$ must be empty, thus $\alpha$ connects to every other root.

**Proposition:** an isomorphism between two irreducible root systems must be conformal, that is, it must scale the metrics by a constant factor.

*Proof:* To prove that a linear map $\phi$ is conformal, it is enough to check this on a generating set of vectors. By definition, an isomorphism $\phi$ preserves the Killing number between any two roots $\alpha$ and $\beta$: $< \phi(\alpha),\, \phi(\beta) > \,=\, < \alpha,\, \beta >$. In terms of the inner product $(\cdot,\cdot)$ this is

$\displaystyle \frac{2(\phi(\alpha),\phi(\beta))}{(\phi(\beta),\phi(\beta))}
= \frac{2(\alpha,\beta)}{(\beta,\beta)}
$

and thus

$ (\phi(\alpha),\phi(\beta)) = c_\beta\, (\alpha,\beta) \qquad$ where $
c_\beta = \displaystyle \frac{(\beta,\beta)}{(\phi(\beta),\phi(\beta))}
$

We can exchange $\alpha$ and $\beta$ in the argument above to get
$ (\phi(\beta),\phi(\alpha)) = c_\alpha\, (\beta,\alpha) $. Since the inner product is symmetric we get

$ (\phi(\alpha),\phi(\beta)) = c_\alpha\, (\alpha,\beta) \qquad$ where $
c_\alpha = \displaystyle \frac{(\alpha,\alpha)}{(\phi(\alpha),\phi(\alpha))}
$

From the two previous equations we have that $c_\alpha = c_\beta$ whenever $(\alpha,\beta) \neq 0$.

By the irreducibility and the previous Lemma we get that $c_\alpha$ is the same for every root, and thus this is the constant scaling factor.

**Corollary:** an automorphism of an irreducible root system must be an isometry.

*Proof:* By the previous Proposition, this automorphism $\phi$ must be conformal. Let $c>0$ be the conformal scaling factor and let $L$ be the norm of the longest root in $\Phi$. Then $\sqrt{c}L$ is the norm of the longest root in $\phi(\Phi)$. Since $\phi(\Phi) = \Phi$, we have that $\sqrt{c}L = L$ so that $\sqrt{c}=1$ and thus $c=1$ and $\phi$ is an isometry.