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?



Do mutually dual finite vector spaces have the same orbit cardinalities under a linear group action?
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 $3,3,2,1$. 


