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Let $G$ be a finite group acting linearly on a finite dimensional vector space $V$ over a finite field. By Burnside's lemma, $$ |V/G| = \frac 1{|G|} \sum_{g\in G} q^{\dim(ker(g - I))}. $$ Since $g-I$ and its dual map $g^*-I$ have kernels of the same dimension, it follows that $|V/G|=|V^*/G|$.

The above argument shows that a vector space and its dual have the same number of orbits under the action of linear group. Can it happen that the cardinalities of the orbits are different?

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    $\begingroup$ One can also prove that the number of orbits are the same using the Fourier transform. It gives a canonical isomorphism $Fun(V) \to Fun(V^*)$ and hence $Fun_G(V) \to Fun_G(V^*)$. The dimensions of both sides is the number of $G$-orbits. $\endgroup$
    – Geordie Williamson
    Commented Apr 15, 2014 at 13:25

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Yes. In particular, it can happen that $V$ has non-zero fixed points, but $V^*$ doesn't.

For example, let $G$ be the symmetric group of degree 3 acting in the obvious way on the set $\{e_1,e_2,e_3\}$, and let $W$ be the corresponding permutation module over the field of 3 elements. Let $V$ be the submodule spanned by $e_1-e_2$ and $e_1-e_3$. Then $V$ has orbits of lengths $6,1,1,1$, but $V^*$ has orbits of lengths $3,3,2,1$.

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