Consider $\mathbb{Q}(\zeta_8) / \mathbb{Q}$. I know that this has a Galois group isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. Therefore, our automorphisms better have the structure that they square to give the identity. Furthermore, let $\{\alpha_1, \alpha_2,\alpha_3,\alpha_4\} := \{\zeta_8, \zeta_8^3, \zeta_8^5,\zeta_8^7\}$ be an enumeration of the roots
This gives us the automorphisms
\begin{align*} \sigma_1 = ID,\hspace{0.2cm} \sigma_2: \begin{cases} \alpha_1 \mapsto \alpha_2\\ \alpha_2 \mapsto \alpha_1\\ \alpha_3 \mapsto \alpha_4\\ \alpha_4 \mapsto \alpha_3 \end{cases} \sigma_3: \begin{cases} \alpha_1 \mapsto \alpha_1\\ \alpha_2 \mapsto \alpha_2\\ \alpha_3 \mapsto \alpha_4\\ \alpha_4 \mapsto \alpha_3 \end{cases} \sigma_4: \begin{cases} \alpha_1 \mapsto \alpha_2\\ \alpha_2 \mapsto \alpha_1\\ \alpha_3 \mapsto \alpha_3\\ \alpha_4 \mapsto \alpha_4 \end{cases} \end{align*}
We have found the correct automorphisms since they square to give the identity, preserving the structure of the Klein four group.
The next thing I want to know is the fixed fields corresponding to subgroups of the Galois group. As far as I know, there are 3 non-trivial subgroups \begin{align*} H_1=\langle\sigma_2\rangle,\hspace{0.2cm} H_2=\langle\sigma_3\rangle,\hspace{0.2cm} H_3=\langle\sigma_4\rangle \end{align*} Which infact are all isomorphic to $\mathbb{Z}/2$.
As far as I know finding these fixed fields means taking the extension field $E:= \mathbb{Q}(\zeta_8)$ and running the permutations over elements in it and checking if they're preserved. The only ones which stick out are the sums in each permutation. These are the elements
\begin{align*} E^{H_1} &= \{\alpha \in E \hspace{0.2cm}\mid\sigma\alpha = \alpha \hspace{0.2cm}\forall\sigma\in H_1\}\\ &= \mathbb{Q}(\zeta_8 + \zeta_8^3, \zeta_8^5 + \zeta_8^7) \end{align*}
And I gather that for $H_2$, $H_3$ the process is similar. These fixed fields are the elements in bijection with the subgroups of the Galois group.
My questions are, have I fully determined the fixed field? How do I know? More over, does this reduce and give us something more familiar?
Thanks!
EDIT: Due to the help of several posters below, I was able to determine that the automorphisms are actually given by
\begin{align*} \sigma_1 = ID,\hspace{0.2cm} \sigma_2: \zeta_8^k \mapsto (\zeta_8^3)^k,\hspace{0.2cm} \sigma_3: \zeta_8^k \mapsto (\zeta_8^5)^k,\hspace{0.2cm} \sigma_3: \zeta_8^k \mapsto (\zeta_8^7)^k \end{align*}
Furthermore that the subgroups of the Galois group are still given as above.
And the fixed fields \begin{align*} \mathbb{Q}(\zeta_8)^{H_1} &= \mathbb{Q}(\zeta_8 + \zeta_8^3, \zeta_8^5 + \zeta_8^7) = \mathbb{Q}(i\sqrt{2})\\ \mathbb{Q}(\zeta_8)^{H_2} &= \mathbb{Q}(\zeta_8 + \zeta_8^5, \zeta_8^5 + \zeta_8^7) = \mathbb{Q}\\ \mathbb{Q}(\zeta_8)^{H_3} &= \mathbb{Q}(\zeta_8 + \zeta_8^7, \zeta_8^3 + \zeta_8^5) = \mathbb{Q}(\sqrt{2})\\ \end{align*}
These were found by actually drawing the 8 roots of unity in the complex plane and using the fact that $$\zeta_8 = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$$