Question: Prove that $\text{Aut}(\mathbb{Q}(i, \sqrt{2})) \cong V_{4}$, where $V_{4}$ is the Klein 4 group.
Attempt: We know that under an automorphism $\sigma \in \text{Aut}(\mathbb{Q}(i, \sqrt{2}))$ the automorphism fixes the prime field $\mathbb{Q}$. Therefore, the image of such automorphisms is entirely determined by the image of $i$ and $\sqrt{2}$.
This is where my concern is. I am pretty sure that the only possibilities for automorphisms would be the identity map, the map where $i \mapsto i$ and $\sqrt{2} \mapsto -\sqrt{2}$, the map where $i \mapsto -i$ and $\sqrt{2} \mapsto \sqrt{2}$ and finally the map $i \mapsto -i$ and $\sqrt{2} \mapsto -\sqrt{2}$. One possibility I could think of is to consider the field extension $\mathbb{Q}(i, \sqrt{2})$ as some splitting field of a polynomial with the corresponding four roots, meaning that a root would also have to map to another root. However, I am unsure if this is the right way to think about the automorphisms. If anybody could explain to me if this is the right way to think about the automorphisms, I would gladly appreciate that.
Your determination of the automorphisms is correct. Instead of trying to find and work with a polynomial for which $\newcommand{\Q}{\mathbb{Q}}\Q(\sqrt{2}, i)$ is a splitting field, here's a standard approach using the tower $\Q(i, \sqrt{2}) / \Q(\sqrt{2}) / \Q$: The extension $\Q(\sqrt{2}) / \Q$ is Galois of degree 2, and the nontrivial element of $\newcommand{\Gal}{\operatorname{Gal}}\Gal(\Q(\sqrt{2}) / \Q)$ is the automorphism taking $\sqrt{2}$ to $-\sqrt{2}$. Now $\Q(i, \sqrt{2}) / \Q(\sqrt{2})$ is also quadratic since $i \notin \Q(\sqrt{2})$, thus by the extension theorem $\Gal(\Q(i, \sqrt{2}) / \Q(\sqrt{2}))$ consists of extending the two automorphisms of $\Gal(\Q(\sqrt{2}) / \Q)$ by either $i \mapsto i$ or $i \mapsto -i$, and you end up with the description you gave.