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Following my previous question (not necessary to refer to), let the number of automorphisms of an $n$-dimensional vector space be $k$. Then am I right in thinking that the number of $m$-dimensional subspaces is simply ${n\choose m}k$? Thanks again.

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up vote 1 down vote accepted

This is obviously false: take $m=n$.

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Unless I misunderstand, taking $m=n$ gives $k$ again, which was posited to be the number of automorphisms of the $n$-dimensional subspace; this would not be a counter-example. However, taking $m=0$.... – Shaun Ault Sep 14 '11 at 12:30
@Shaun: an $n$-dimensional space has precisely one $n$-dimensional subspace. – Chris Eagle Sep 14 '11 at 12:31

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