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Let $q=p^f$ be a prime power, $V$ is a $n$-dimensional vector space over $GF(q)$ and $G=GL(n,q)=GL(V)$. Is every transitive permutation representation $\rho$ of $G$ on $q^n-1$ points isomorphic to the natural action of $\tau\in G$ on $V\setminus\{0\}$?

Thanks in advance!

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    Yes sorry you are right! Ignore all my comments about 2. I believe the answer is no. I will try and send some examples later.2012-11-09

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Apologies for all of my incorrect comments about Question 2. The correct answer is yes for $q=2$, but no otherwise.

A transitive permutation representation of $G$ of degree $q^n-1$ corresponds to a subgroup of $G$ of index $q^n-1$. The stabilizer $H$ of a 1-dimensional subspace of $V$ is a maximal subgroup of $G$ of index $(q^n-1)/(q-1)$. It has a normal elementary abelian subgroup $N$ of order $q^{n-1}$, and $H/N$ is isomorphic to a direct product $C_{q-1} \times {\rm GL}_{n-1}(q)$. This is easy to see directly.

Thje complete inverse image in $H$ of the direct factor ${\rm GL}_{n-1}(q)$ is the stabilizer of a nonzero vector in $V$, and so corresponds to the natural action on vectors. But for $q>2$ there are several other subgroups of $H$ with the same index $q-1$, such as the inverse image in $H$ of $C_{q-1} \times {\rm SL}_{n-1}(q)$. Since these are not isomorphic to the stabilizer of a vector, their corresponding permutation representations are not isomorphic to the natural one.