Galois group of $x^4-2$

I am trying to explicitly compute the Galois group of $x^4-2$ over $\mathbb{Q}$. I found that the resolvent polynomial is reducible and the order of the Galois group is $8$ using the splitting field $K=\mathbb{Q}(2^{1/4}, i)$. Hence I need to find 8 automorphisms. Thus do I just map elements of the same order to each other that fix the base field $\mathbb{Q}$ $2^{1/4}$ to $i2^{1/4}$

You can see that $i^n\sqrt[4]{2},\, 0\le n\le 3$ are all the roots, so that

$$K=\Bbb Q(i,\sqrt[4]{2})$$

is the splitting field for the polynomial. Since $L=\Bbb Q(\sqrt[4]{2})$ is real of degree $4$, we see that $K$ is a proper extension of $L$, and since $[\Bbb Q(i):\Bbb Q]=2$ we see the total degree of the extension is $2\cdot 4=8$. But then we have that $\operatorname{Gal}\left(K/\Bbb Q\right)\le S_4$ is a subgroup of $S_4$ of order $8$. This implies it is a Sylow-$2$ subgroup of $S_4$, all of which are isomorphic--by the second Sylow theorem. We know that $D_8$, the dihedral group of order $8$, is such a subgroup, so that gives the isomorphism type.

But then you know what to look for as explicit representations go, you note that relative to the ordering

$$\alpha_j=i^j\sqrt[4]{2}, 1\le j\le 4$$

we have the $4$-cycle $(1234)$ given by the automorphism

$$\begin{cases}\sqrt[4]{2}\mapsto i\sqrt[4]{2} \\ i\mapsto i\end{cases}$$

which is enough to totally determine it, since those are generators of the extension. Clearly also

$$\begin{cases}\sqrt[4]{2}\mapsto \sqrt[4]{2} \\ i\mapsto -i\end{cases}$$

is represented by the transposition $(13)$, and these two generate the group, so give you everything you need for a fully explicit description.

You should note that the Galois group is isomorphic to a subgroup of $S_4$, since any element of the Galois group will be determined by how it permutes the four roots of $x^4-2$. The order 8 subgroups of $S_4$ are isomorphic to the dihedral group with 8 elements.

If you want to know exactly what the eight maps are, there are four choices as to where $\sqrt[4]{2}$ is sent under some element of the Galois group. That tells you where the root $-\sqrt[4]{2}$ is sent as well, and then you have two choices as to where $i\sqrt[4]{2}$ is sent.