Smooth morphism to homogeneous spaces and fibers Let $f:X \to Y$ be a smooth morphism between projective varieties. Suppose $Y$ is a homogeneous space. Under what additional condition on $f$, can we conclude that every fibers of $f$ are isomorphic?
 A: For your particular case, there is a sufficient hypothesis that the morphism is étale locally a product.  Let $k$ be a field.  Let $G$ be a smooth $k$-group scheme.  Let $$\Psi_X:G\times_{\text{Spec}(k)} X \to X\times_{\text{Spec}(k)} X, \ \ \Psi_X(g,x) = (g\cdot x,x), $$
$$\Psi_Y:G\times_{\text{Spec}(k)} Y \to Y\times_{\text{Spec}(k)} Y, \ \ \Psi_Y(g,y) = (g\cdot y,y)$$
be $k$-actions of $G$ on $X$, resp. $Y$.  Assume that $f$ is $G$-equivariant, $f(g\cdot x) = g\cdot f(x)$.  Finally, assume that the action $\Psi_Y$ is smooth and surjective, i.e., $Y$ is a $G$-homogeneous space whose stabilizer subgroup schemes are $k$-smooth.  Then the morphism $f$ is étale locally a product.  
It suffices to prove this after étale base change, thus assume that $Y$ has a $k$-point $y$.  Then $\Psi_Y$ induces a smooth $k$-morphism, $$\psi_{Y,y}:G\to Y,\ \ \psi_{Y,y}(g) = g\cdot y.$$  Denote $X\times_{f,Y,y} \text{Spec}(k)$ by $X_y$.  Then there is a commutative diagram.
$$ \begin{array}{ccc}
G\times_{\text{Spec}(k)} X_y & \xrightarrow{\text{pr}_1\circ \Psi_X} & Y \\
\ \ \ \ \ \ \downarrow{\text{pr}_G} & & \downarrow{f} \\
G & \xrightarrow{\psi_{Y,y}} & X
\end{array}$$
It is not too hard to check that this is actually a Cartesian diagram.  Thus, after the smooth base change $\psi_{Y,y}$, the morphism $f$ becomes a product.  Of course there are étale local sections of $\psi_{Y,y}$.  Thus, after étale base change of $Y$, the morphism $f$ becomes a product. 
