Given that a finite vector space $V = \operatorname{GF}(p)^n$ corresponds to the finite field $F = \operatorname{GF}(p^n)$, I'm wondering about how the multiplicative subgroup of $F$, $F^*$, acts on subspaces of $V$. Since $F$ is a field, we know that any element of $F^*$ is a linear automorphism of V, and is in $\operatorname{GL}(V)$. It therefore maps $k$-dimensional subspaces to $k$-dimensional subspaces. Further, we know that $F^*$ is the cyclic group of order $p^n - 1$. However, there are a couple things I'm wondering:
1) When will the action, for a given $p^n$ and $k$, be semi-regular: i.e. when will there be at most one $x \in F^*$ that maps between two $k$-dimensional subspaces $U_1$ and $U_2$.
2) What will the orbits under this action look like? Particularly, when will several subspaces in the orbit of a given $t$-dimensional space lie in a $t+1$-dimensional subspace, for example when will several 3-space in the same orbit lie in a single 4-space. I know that in the highly specific case where $p=2$ and $gcd(n,6)=1$, there are a number of 3-spaces (projective planes) containing three 2-spaces (projective lines) in the same orbit equal to the number of 2-spaces (projective lines) in the vector space. But do we know that, in general, an orbit will intersect like this?
I especially care about the case where $p=2$: if anyone knows anything about the action of $\operatorname{GF}(2^n)^*$ on its underlying vector space, or any books / articles that talk about this stuff, it would be tremendously appreciated.
Thank you!
(The question ultimately relates to a project where I am looking for designs over a finite field, and hence am trying to find collections of subspaces with certain regularities. Orbits under the multiplicative group seem very useful. Some related ideas come from: Thomas, Simon. "Designs over finite fields." Geometriae Dedicata 24.2 (1987): 237-242.)
The correspondence through the use of the Field Reduction map will give a canonical way to view $\mathbb{F}_{q^n}$ as $\mathbb{F}_{q}^{n}$.
1) $\mathbb{F}_{q^n}^{*}$ will not act semiregularly on the $k$-spaces of $PG(n-1,q)$ for odd $q$, because the elements in the base field $\mathbb{F}_{q}$ will act as the identity. Although for $q=2$ this is okay. In general, you can consider a subgroup $\langle \mu \rangle$, where $\mu \in \mathbb{F}_{q^{n}}^*$ and ask if this group acts semi-regularly.
For q=2, you are in a very special case; $\mathbb{F}_{q^n}^{*}$ acts as a Singer cycle on the space, a cyclic group acting sharply transitive on the points. The action on other subspaces has been studied in
D. Glynn, On a set of lines of PG(3,q) corresponding to a maximal cap contained in the Klein quadric of PG(5,q), Geom. Dedicata 26 (1988) #3, 273-280. (lines)
K. Drudge, On the orbits of Singer groups and their subgroups, Elec. J. of Comb. 9 (2002)