Let $L/F$ be a finite galois extension of fields, with galois group $\Gamma$. Let $X$ be a variety over $F$ such that $X_L \cong \mathbb{P}^1_L$ over $L$, corresponding to a cohomology class $\alpha \in H^1(\Gamma,\text{PGL}_{2,L})$; $\alpha$ determines an inner $F$-form $G$ of $\text{GL}_{2}$ by means of the map
$$H^1(\Gamma,\text{PGL}_{2,L}) \rightarrow H^1(\Gamma,\text{Aut}(\text{GL}_{2,L})).$$
Can $X$ be regarded as a homogeneous space for $G$ somehow?
It is not an accident that twisted forms of $P^1$ and twisted forms of $GL_2$ are classified by the same data. This is nicely explained in these slides: the cohomology class $\alpha$ determines a quaternion algebra $B$ over $F$ (split over $L$), and both $G$ and $X$ are naturally determined by $B$: the group $G$ is just the unit group $B^\times$, while $X$ is the variety of 2-dimensional right ideals of $B$.
I don't think that $X$ will be a homogenous space for $G$, though; if $B$ is not the split algebra, then $G$ is anisotropic (has no nontrivial parabolic subgroups) and hence cannot act transitively on any nontrivial projective variety.EDIT. I thought about this again and I realised that the above paragraph is wrong. If $E/F$ is any extension, then the stabiliser of any $x \in X(E)$ will be a parabolic subgroup of $G$ defined over $E$. If $B$ is not the split algebra, then we know that $G$ doesn't have any parabolic subgroups defined over $F$; but that isn't a contradiction, because $X(F)$ is empty!
If one uses the description of $X$ as the variety of 2-dimensional right ideals in $B$, then it is actually obvious that $B^\times$ acts on $X$ (by left multiplication of ideals). Over a suitable extension (such as $L$) this just becomes the usual action of $GL_2 / L$ on $\mathbf{P}^1_L$, so it has to be transitive. So $X$ is indeed a homogenous space for $G$.