 Solution

Yes. Suppose $g$ maps the first flag to the second and has determinant $\delta \ne 0$. 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.