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I think I have a proof of the following elementary lemma (although I only need the case in which the two flags are "in general position", i.e., $F^d \cap G^i$ is minimal given the dimensions of the spaces):

Let $V$ be a finite-dimensional vector space, and $F^{\bullet}$, $G^{\bullet}$ two decreasing partial flags on $V$. Suppose that for every $d$, a linear map $$ \phi_d \in \operatorname{End}(F^d / F^{d+1}) $$ is given that respects the decreasing partial flag on $F^d / F^{d+1}$ defined by $$\newcommand{\dpunct}{\;\text} G^i_d = \frac{G^i \cap F^d}{G^i \cap F^{d+1}}\dpunct. $$ Then there exists a linear map $ \phi \colon V \to V $ respecting both $F^{\bullet}$ and $G^{\bullet}$ such that for every $d$, the induced endomorphism of $F^d / F^{d+1}$ is precisely $\phi_d$.

I'm sure this is in some sense standard. My proof certainly does not feel like something that belongs in a research paper. Does anyone know a standard reference either for this fact, or for other standard facts from which this lemma follows trivially?

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$V\cong\bigoplus_d F^d/F^{d+1}$ and the $G^i_d$ encode the flag $G^*$ in this representation. From this the result follows. I would not even formulate it as a lemma.

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    $\begingroup$ That is an excellent suggestion. Unfortunately, unless the splitting $V \cong \bigoplus_d F^d/F^{d+1}$ is chosen carefully, it is not true that the G^i_d encode the flag $G^*$. For instance, if $F^0 \supset F^1 \supset F^2$ and $G^0 \supset G^1 \supset G^2$ are complete flags of $\Bbbk^2$ satisfying $G^1 \cap F^1 = 0$, then the $G^i_d$ do not encode $G^*$ unless the splitting $F^1 \oplus F^0/F^1 \cong V$ is chosen such that the image of $F^0 / F^1$ is precisely $G^1$. $\endgroup$
    – Charles Staats
    Commented Aug 19, 2013 at 16:21

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