My question in general is this: given a number field $K$ (not necessarily Galois), how can the Galois group of the Galois closure $\text{Gal}(\widetilde{K}/\mathbb{Q})$ act on embeddings of $K$ into $\overline{\mathbb{Q}}$? Furthermore, how can one say such an action is isomorphic to $G$ for some group $G$? I am basically referring to the first paragraph of the paper on "The density of discriminants of $S_3$-sextic number fields" by Bhargava and Wood.
Here is a link to the paper: https://www.ams.org/journals/proc/2008-136-05/S0002-9939-07-09171-X/S0002-9939-07-09171-X.pdf
Also, is this the same thing as saying $\text{Gal}(\widetilde{K}/\mathbb{Q}) \cong G$? If so, how? I appreciate any help I can get (a simple example like $K = \mathbb{Q}(\sqrt[3]{2})$ would be really helpful).
Thank you.
Fix an embedding $\iota: \widetilde K \hookrightarrow\overline{\mathbb Q}$. Then every embedding $\sigma : K \hookrightarrow \overline{\mathbb Q}$ has image lying in $\iota(\widetilde K)$. Now $\operatorname{Gal}(\widetilde K / \mathbb Q)$ acts on $\operatorname{Hom}(K, \overline{\mathbb Q})$ by composition on the right: Define $$\forall \tau \in \operatorname{Gal}(\widetilde K / \mathbb Q), \sigma \in \operatorname{Hom}(K, \overline{\mathbb Q}) : \tau * \sigma := \iota \circ \tau \circ \iota^{-1} \circ \sigma \,.$$ This gives a group action $\operatorname{Gal}(\widetilde K / \mathbb Q) \curvearrowright \operatorname{Hom}(K, \overline{\mathbb Q})$. It is faithful.
Note: this group action depends on the choice of embedding $\iota$! There is no 'canonical' way to define such an action. Another choice of $\iota$ gives rise to a group action that is equal to this one up to conjugation by an element of $\operatorname{Gal}(\widetilde K / \mathbb Q)$. But the image of the permutation representation does not depend on $\iota$.
What I assume is meant by isomorphic in the paper is that the image of the permutation representation $\operatorname{Gal}(\widetilde K / \mathbb Q) \to \operatorname{Sym}(\operatorname{Hom}(K, \overline{\mathbb Q}))$ is isomorphic to $G$ as a permutation group. That is, when $K$ has degree $n$, that there is a bijection $\{1, \ldots, n\} \to \operatorname{Hom}(K, \overline{\mathbb Q})$ under which $G$ has the same image as $\operatorname{Gal}(\widetilde K / \mathbb Q)$.
Example: Take $K = \mathbb Q(\sqrt 2, \sqrt 3)$ so that $\widetilde K = K$. We have $|\operatorname{Hom}(K, \overline{\mathbb Q})| = 4$ and the image of $\operatorname{Gal}(K/\mathbb Q)$ consists of the permutations $(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3) \in S_4$ with $\delta, \epsilon \in \{0, 1\}$, provided we identify $\operatorname{Hom}(K, \overline{\mathbb Q}) \cong \{1, 2, 3, 4\}$ in the right way.
As an abstract group, we also have $\operatorname{Gal}(K/\mathbb Q) \cong \{(), (1, 2), (3, 4), (1, 2)(3, 4)\}$, simply because the latter is also isomorphic to $(\mathbb Z/2 \mathbb Z)^2$. But they are not isomorphic as permutation groups. I think this type of abstract isomorphism is not allowed in the paper.