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 $(\mathbb{Z}/2\mathbb{Z})^2$, 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.)