Given the $20^{th}$ primitive root of unity $\omega$, we know that it's splitting field is $\Bbb Q(\omega)$. Furthermore we know that it's a Galois extension and that it's minimal polynomial over $\Bbb Q$ is $x^8-x^6+x^4-x^2+1$, which had degree 8 , so the Galois group has degree 8. We can also see this from the fact that the roots of the minimal polynomial are exactly those powers of $\omega$ which are coprime to $20$. From this second fact We can deduce that the Galois group must be isomorphic to $\Bbb U_{20} \cong\Bbb U_5\times\Bbb U_4 \cong C_4\times C_2$.
This implies that the structure of the group is going to consist of transpositions and 4 cycles.
We in fact know that each automorphism will be of the form $\phi_k:\omega\rightarrow \omega ^k$ where $k$ is coprime to, and less than, $n $.
So we can list the automorphisms explicitly as :
$e:\omega \rightarrow \omega$
$\phi_2:\omega \rightarrow \omega^3$
$\phi_3:\omega \rightarrow \omega^9$
$\phi_4:\omega \rightarrow \omega^7$
$\phi_5:\omega \rightarrow \omega^{11}$
$\phi_1:\omega \rightarrow \omega^{13}$
$\phi_6:\omega \rightarrow \omega^{19}$
$\phi_7:\omega \rightarrow \omega^{17}$
These ,respectively, correspond exactly to how the elements of $D_8=\langle r,s|r^4=s^2=1 ,s^{-1}rs=r^{-1} \rangle$ would act. Which leads me to believe that the group is isomorphic to $D_8$. But here https://en.wikipedia.org/wiki/Semidirect_product in the examples of the outer semi direct product section it says that $D_{2n}$ isomorphic to a semidirect product of the cyclic groups $C_n$ and $C_2$, while in this case we only have a direct product? What is happening here , the structures are identical I believe, yet it seems we don't have the conditions for isomorphism? I know I must be wrong about one of those things.