I like this question because it speaks to a natural misunderstanding: why can't we always send any root to any other one? I think the best way to absorb this idea is by example.
Example 1: Let $K = \mathbb{Q}[\sqrt{2},\sqrt{3}]$ and notice that $K/\mathbb{Q}$ has degree four and is the splitting field of the polynomial $(x^2 - 2)(x^2 - 3)$. Thus, $K/\mathbb{Q}$ is a Galois extension of degree four so, as you mention, $| \mathrm{Gal}(K/\mathbb{Q}) | = 4$. In fact, this Galois group has generators $\sqrt{2} \mapsto -\sqrt{2}$ and $\sqrt{3} \mapsto -\sqrt{3}$ and is thus isomorphic to the Klein four group. Now to your question: why isn't the Galois group just $S_4$? Here two ways to think about it:
One way is to leverage the fact that a $\mathbb{Q}$-automorphism must send roots of polynomials over $\mathbb{Q}$ to other roots of the same polynomial. In our example: take $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$ and consider the polynomial $x^2 - 2$. Then the fact that $\sigma$ is an automorphism fixing $\mathbb{Q}$ implies $\sigma(x^2 - 2) = \sigma(x)^2 - 2$, which in turn implies that $\sigma(x)$ is itself a root of $x^2 - 2$. That is, $\sigma(x) = \sqrt{2}$ or $\sigma(x) = -\sqrt{2}$. There are no other options. For $\mathrm{Gal}(K/\mathbb{Q})$ to equal $S_4$, we would need automorphisms which can permute any pair of roots, but that cannot happen here!
A less precise way of thinking about this is to ponder automorphisms conceptually: automorphisms --- as isomorphisms --- should preserve the structure of the field. In other words, once you apply an automorphism $\sigma \in \mathrm{Gal}(K/\mathbb{Q})$, you should wind up with a field $\sigma(K)$ algebraically indistinguishable from $K$. So $\sqrt{2}$ and $\sigma(\sqrt{2})$ should have exactly the same algebraic properties. This works for $\sqrt{2}$ and $-\sqrt{2}$ (the ordering of the real numbers and the fact that $-\sqrt{2} < \sqrt{2}$ is an analytic, not algebraic, concern), but does not work for $\sqrt{2}$ and $\sqrt{3}$ because --- for example --- $(\sqrt{2})^2 = 2 \neq 3 = (\sqrt{3})^2$ (if $2 = 3$ then $0 = 1$ and we're dealing with the "zero ring").
This example teaches the following lesson: the Galois group is all about preserving the symmetry of a field, but the amount of symmetry is necessarily limited. Elements of a field will often have subtle/nuanced interactions, and the Galois group picks up on that! If $\mathrm{Gal}(K/\mathbb{Q})$ were always a full symmetric group, then that would suggest that we always have a maximum amount of symmetry (but we don't).
Exercise: Analyze the situation for the field $K = \mathbb{Q}[e^{2 \pi i/4}]$.
Example 2: Let $\zeta_3 = e^{2\pi i/3}$ be a primitive third root of unity and let $K = \mathbb{Q}[2^{1/3}, \zeta_3]$, an extension of degree six (take this on faith for now and then prove it later once you understand this example). Then $\mathrm{Aut}(K/\mathbb{Q})$ contains an element $\sigma$ which sends $\zeta_3$ to $\zeta_3^2$ and fixes everything unrelated to $\zeta_3$ ($\sigma$ is also known as complex conjugation); to make perfectly clear what $\sigma$ does, here are some examples of its action:
- $\sigma(\zeta_3^2) = \sigma(\zeta_3)^2 = \zeta_3^4 = \zeta_3$,
- $\sigma(2^{1/3}) = 2^{1/3}$,
- $\sigma(\zeta_3 + 2^{1/3}) = \zeta_3^2 + 2^{1/3}$, and
- $\sigma(q) = q$ for all $q \in \mathbb{Q}$.
Note that the first bullet implies that $\sigma^2$ is the identity map. (You should pause here to prove that $\sigma$ indeed defines a $\mathbb{Q}$-automorphism of $K$.) Similarly, $\mathrm{Aut}(K/\mathbb{Q})$ contains an element $\tau$ which sends $2^{1/3}$ to $\zeta_3 2^{1/3}$ and fixes everything unrelated to $2^{1/3}$ (as before, prove that $\tau$ is an automorphism, compute some examples of its action, and show that $\tau^3$ is the identity.) Then $\sigma$ and $\tau$ generate all of $\mathrm{Aut}(K/\mathbb{Q})$ (why?) so $\mathrm{Aut}(K/\mathbb{Q}) = < \sigma, \tau >$. We thus have a natural isomorphism of groups
\begin{align*}
\mathrm{Aut}(K/\mathbb{Q}) &\to S_3 \\
\sigma &\mapsto (2\, 3) \\
\tau &\mapsto (1\, 2\, 3)
\end{align*}
so in fact $K/\mathbb{Q}$ is a Galois extension with Galois group $S_3$. How does this example relate to the discussion of permuting roots of a polynomial? Without going into too many details, we have $K = \mathbb{Q}[2^{1/3}, \zeta_3] = \mathbb{Q}[2^{1/3} + \zeta_3]$ where $2^{1/3} + \zeta_3$ has minimum polynomial $$p(x) = x^6 + 3 x^5 + 6 x^4 + 3 x^3 + 9 x + 9$$ over $\mathbb{Q}$. The roots of $p(x)$ are
- $2^{1/3} + \zeta_3$
- $\zeta_3 2^{1/3} + \zeta_3$
- $\zeta_3^2 2^{1/3} + \zeta_3$
- $2^{1/3} + \zeta_3^2$
- $\zeta_3 2^{1/3} + \zeta_3^2$
- $\zeta_3^2 2^{1/3} + \zeta_3^2$
so that $\sigma$ and $\tau$ together shuffle around all roots of $p(x)$! In some sense, this example therefore has the maximum possible amount of symmetry.
Now, to the extensions 1) and 2) in your question: as Jyrki notes in their comment on your question, neither extension is Galois (over $\mathbb{Q}$). I hope that the examples I've explained help you to see why.
EDIT: As pointed out in the comments, I neglected to address the construction of a Galois group isomorphic to $\mathbb{Z}/n!\mathbb{Z}$, the cyclic group of order $n!$. I think the easiest way to do this is to apply the fundamental theorem of Galois theory to a cyclic Galois extension. Indeed, let $n!$ divide $d$, where $d$ is the degree of some cyclic Galois extension $L/K$. Then there is a cyclic subgroup of $\mathrm{Gal}(L/K)$ of degree $n!$ and the fixed field of this subgroup will be the desired Galois extension (note that we are using that cyclic groups are abelian here to get that the intermediate extension is Galois). There are many options for the cyclic Galois extension $L/K$ (e.g. extensions of finite fields, whose Galois groups are comparatively straightforward).
Subgroups $H$ of $Gal(E/F)$ correspond to intermediate fields $K$ where $H = Gal(E/K)$ Therefore, if we apply it to $F = \mathbb{Q}$, then all we can conclude is that $H$ is the Galois group of some field extension $E/K$, neither of which are required to be $\mathbb{Q}$.
To solve the inverse Galois problem, it is sufficient to show that every finite group is a quotient of $S_n$, not a subgroup.
Best Answer
Yes, every finite group is the Galois group of some extension. Consider the field $K = \mathbb{Q}(e_1, \dots e_n)$ and its extension $L = \mathbb{Q}(x_1, \dots x_n)$ where the $x_i$ satisfy
$$\prod (t - x_i) = \sum (-1)^i e_i t^{n-i}$$
where $e_0 = 1$, so the $e_i$ are the elementary symmetric polynomials in the $x_i$. Then $L$ is a finite Galois extension of $K$ with Galois group $S_n$ (exercise). Since every finite group $G$ embeds in some $S_n$, we get finite Galois extensions $L$ of fields $L^G$ with Galois group $G$.
The inverse Galois problem is hard not because we don't know enough about finite groups but because we don't know enough about the Galois theory of $\mathbb{Q}$. It's equivalent to showing that every finite group is a quotient of the absolute Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, and this is just hard. The abelian case is much easier because abelian quotients of the Galois group are handled abstractly by class field theory and concretely by the Kronecker-Weber theorem, but "nonabelian class field theory" is much harder, not completely understood, and very much the subject of active research.