\b[Yes.] Suppose $g$ maps the first flag to the second and has <determinant> $\delta\ne0$. Relative to an adapted <basis> of the first flag, scaling its first <basis> vector by $\delta^{-1}$ and fixing all others defines a transformation $h$ preserving that flag, with <determinant> $\delta^{-1}$. Then $gh$ still maps the first flag to the second and has <determinant> one. Thus the special linear <subgroup> acts transitively too.

Solution

Past exam of the options II course of the University of Oxford / 2008 printed variant / A / A2 / b / ii / Solution

 Solution (source code)