 Solution

Given two flags of the same dimensions, choose adapted ordered bases $u_1,\dots ,u_4$ and $v_1,\dots ,v_4$, so their first $d_i$ vectors span their respective $L_i$. The invertible map $A$ sending each $u_j$ to $v_j$ sends the first flag to the second. Let $\delta =\det A\ne 0$. Replace the last target basis vector by ${\delta }^{-1}{v}_4$; this does not alter any subspace in its flag. The corresponding map $g$ has determinant one and still sends every member of the first flag to the corresponding member of the second.

Thus the special linear group acts transitively on every fixed-type flag manifold:The adapted-basis construction is the needed extension of transitivity on nonzero vectors. For an arbitrary subgroup, vector transitivity alone would not justify transitivity on all flags; here the explicit determinant correction proves the stronger claim for the special linear group.