Conjugacy classes of maximal tori in PGL2: only 2? 
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 $K ⊕ 0$.
 A: No. There are many tori when there are many extensions.
Let $L,F$ be quadratic extensions of a field $K.$ Upon choosing a bases for $L$ and $F$ over $K$ we obtain monomorphisms $i_L$ and $i_F$ from $L$ and $F,$ respectively, into the ring $\mathfrak{M}(2,K)$ of 2-by-2 matrices over $K.$ The multiplicative groups $L^{\times}$ and $F^{\times}$ are mapped under this homomorphism into $GL(2,K).$ Let $\pi:GL(2,K) \rightarrow PGL(2,K)$ be the projection map. We claim that if $\pi(i_L(L^{\times}))$ and $\pi(i_F(F^{\times}))$ are conjugate in $PGL(2,K)$ by some element $\overline{A},$ then $L\cong F$.   
Let $A$ be a lift of $\overline{A}$ to $GL(2,K).$ Conjugation by $A$ is an isomorphism from $i_L(L)$ to $i_L(L)^A$ corresponding to a change in the basis chosen for $L/K.$ Thus, applying $i_L$ and acting by $A$ is a monomorphism from $L$ into $GL(2,K)$. Subsequently, we may replace $i_L$ by our original embedding followed by conjugation, and restrict our attention to the case where $\pi(i_L(L^{\times})) = \pi(i_F(F^{\times})).$
Let $\alpha$ be a primitive element of $L/K$ and $\beta \in F$ such that $\pi(i_L(\alpha)) = \pi(i_F(\beta)).$ Then $i_L(\alpha) = a i_F(\beta)$ for some $a\in K.$ It follows that if $p_{\alpha}(X) = X^2 + b_1X + b_0$ is the characteristic polynomial for $i_L(\alpha),$ then $p_{\beta}(X) = X^2 + a b_1X + a^2 b_0 = a^2p_{\alpha}(\frac{1}{a}X)$ is the characteristic polynomial for $i_F(\beta).$ By the Cayley-Hamilton theorem $p_{\alpha}(i_L(\alpha)) = i_L(p_{\alpha}(\alpha))$ and $p_{\beta}(i_F(\alpha)) = i_F(p_{\alpha}(\beta))$ are zero transformations. But $i_L$ and $i_F$ are embeddings, so $p_{\alpha}(\alpha) = 0$ and $p_{\beta}(\beta) = 0.$ It follows, as the degree of $p_{\alpha}$ is equal to [K(\alpha):K], the polynomial $p_{\alpha} = irr(\alpha, K)$ and $irr(\alpha, K)(\frac{\beta}{a^2}) = 0$. We conclude $L = K(\alpha) = K(\alpha/a^2)$ embeds into $F,$ and as the two fields have the same dimension over $K,$ they are in fact isomorphic.
