gr.group theory - Algebraic groups of relative rank 1

Fleshing out the Galois cohomology approach suggested by Brian Conrad leads to a clean answer for all $n$, for all fields $K$ of characteristic not $2$, and for all nondegenerate quadratic forms of rank $n$ over $K$. The answer is exactly what moonface claimed:

Given quadratic forms $q$ and $q'$, the algebraic groups $operatorname{SO}(q)$ and $operatorname{SO}(q')$ are isomorphic if and only if $q$ and $q'$ are similar, i.e., $q$ is equivalent to $lambda q'$ for some $lambda in K^times$.

The key observation is that the homomorphism from $operatorname{O}_n$ to the automorphism group scheme of $operatorname{SO}_n$ giving the conjugation action is surjective for all $n$, and the kernel is $lbrace pm 1 rbrace$ for $n>2$. Then for $n>2$, one has the exact sequence of pointed sets
$$ H^1(K,lbrace pm 1 rbrace) to H^1(K,operatorname{O}_n) to H^1(K,operatorname{bf Aut} operatorname{SO}_n).$$
The first term is $K^times/K^{times 2}$, the second term is the set of equivalence classes of nondegenerate rank $n$ quadratic forms, and the third term is the set of isomorphism classes of $K$-forms of $operatorname{SO}_n$. The sequence (and its twists - remember that we are dealing with nonabelian cohomology) shows that two quadratic forms give rise to the same $K$-form of $operatorname{SO}_n$ if and only if they are similar.

If $n=2$, a similar argument applies, though one can also see everything explicitly: every rank $2$ quadratic form is similar to $x^2-dy^2$ for some $d in K^times$, and the corresponding $operatorname{SO}(q)$ is the "Pell equation torus" $x^2-dy^2=1$; both depend just on the image of $d$ in $K^times/K^{times 2}$. I leave the cases $n=1$ and $n=0$ to those who like to think about such things.

More details: When $n$ is odd, we have $operatorname{O}_n = lbrace pm 1 rbrace cdot operatorname{SO}_n$, and all automorphisms of $operatorname{SO}_n$ are inner.

When $n=2m$ for some $m ge 2$, we have $-1 in operatorname{SO}_n$, so conjugation by an element of $operatorname{O}_n$ outside $operatorname{SO}_n$ gives an outer automorphism of $operatorname{SO}_n$; correspondingly, the $D_m$ Dynkin diagram (interpreted appropriately for small $m$) has an involution.

Why does the rotation of the $D_4$ Dynkin diagram not give an extra outer automorphism when $n=8$? The covering group between the simply connected form $operatorname{Spin}_8$ and the adjoint form $operatorname{PSO}_8$ is $(mathbb{Z}/2mathbb{Z})^2$, so there are three intermediate covers, one of which is $operatorname{SO}_8$, but the rotation permutes these three.

(Thanks to my colleague David Vogan for discussing this with me.)