If $K$ is a field, and $F$, $L$ are two distinct quadratic extensions of $K$, then must the subgroups of $\operatorname{PGL}(2, K)$ defined by $F$ and $L$ be conjugate?
To define the subgroup, identify the field $F$ with a two-dimensional $K$-vector space. The subgroup is the subgroup induced by the action of the nonzero elements of $F$ on this vector space. Different identifications yield conjugate subgroups.
Everything makes good sense when the absolute Galois group of the field is locally cyclic, so that extension fields are extremely unique. However, I'm not sure this makes much sense for a field like $\mathbf{Q}$. What goes wrong with (at least one of) the following arguments:
They should not be conjugate: Surely conjugacy would lift to an isomorphism of $K$-algebras between $F$ and $L$? Surely it simply defines an isomorphism of the groups of units of $F$ and $L$? However, I think not all quadratic fields over $\mathbf{Q}$ have the same group of units (the torsion parts differ at least in $\mathbf{Q}[i]$ and $\mathbf{Q}[ω]$).
They should be conjugate: I believe conjugacy classes of maximal tori correspond in some way to elements of the Weyl group. The Weyl group has two elements, and the identity element corresponds to the subgroup defined by $K$ acting on $K ⊕ 0$.