Let $F$ be a finite field with $p^n$ elements with prime $p > 5$ and $G = S_5$ the symmetric group. Let $l = 60$ be the l.c.m. of the orders of the elements of G and $\theta$ be the primitive $l^{th}$ root of unity over $F$. Define the multiplicative group $$T = \{k\ | \theta \to \theta^k \ \text{ is automorphism of} \ F(\theta) \ \text{over} \ F\}$$ Then what is $|T|$?
What I know is $Gal(F(\theta), F)$ is a cyclic group and as $(60, p) = 1$ image of $Gal(F(\theta), F)$ in $\mathbb{Z}_{60}^*$, mutiplicative group, is $p^n\pmod{ 60}$. So same thing holds for $k$ also. How to think? Please elaborate.