= Solution
Given two flags of the same dimensions, choose adapted ordered bases $u_1,\ldots,u_4$ and $v_1,\ldots,v_4$, 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 $\delta=\det A\ne0$. 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 \b[the special linear group acts transitively on every fixed-type flag manifold]:
$$
\boxed{SL(4,\mathbb C)\cdot[L_1,\ldots,L_m]=F_{d_1,\ldots,d_m}(\mathbb C^4).}
$$
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>.
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