Let $K$ be a field, $V$ an $n-$dimensional $K$-vector space and $q: V \to K$ a quadratic form of Witt index $r$. Let $G:=SO(q)$ denote the special orthogonal group associated to $q$. 
Then $G$ is an algebraic $K$-group of $K-$rank $r$. 
If $q'$ is isometric to $q$, clearly $G':=SO(q')$ is isomorphic to $G$. But this is not necessary for $G$ to be isomorphic to $G'$ (take $n=1$).


My question is: How can I produce a lot of nonisomorphic $SO(q)$'s? 
Here "non-isomorphic" should be interpreted either as abstract groups or as algebraic groups.
I'm particularly interested in the case when the Witt rank is 1. 



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:



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> 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 $\lambda \in K^\times$.
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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 $(\mathbb{Z}/2\mathbb{Z})^2$, 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.)


