Suppose we have a polynomial $P \in \Bbb{Q}[x]$ (a polynomial with rational coefficients) and that $P$ reduces to the product of powers of distinct polynomials $(C_1)^{m_1} (C_2)^{m_2} ... (C_n)^{m_n}$ such that each $C_i$ is an irreducible polynomial. Let $F$ be the splitting field of $P$ that is the "smallest" field over which every root of $P$ is expressible (said another way, the intersection of all fields containing every root of $P$). Let $deg(E)$ be the degree of the polynomial E. Then:
$$ Gal\left( \frac{F}{\Bbb{Q}} \right) = S_{deg(C_1)} \times S_{deg(C_2)} ... S_{deg(C_n)}$$
If we let $ a \le b$ denote $a$ is a subgroup of $b$ then it's easy to show that the left hand is a subgroup of $S_{deg(P)}$.
Observe that every automorphism $\phi \in Gal\left( \frac{F}{\Bbb{Q}} \right)$ corresponds to some permutation of the roots, so it clearly will be contained in the permutation group over the entire set of roots. From here it follows that the closed group of automorphisms forms a subgroup over the permutation group of all the roots.
A general technique of proof can be to establish that
$$ Gal\left( \frac{F}{\Bbb{Q}} \right) \le S_{deg(C_1)} \times S_{deg(C_2)} ... S_{deg(C_n)}$$
$$ Gal\left( \frac{F}{\Bbb{Q}} \right) \ge S_{deg(C_1)} \times S_{deg(C_2)} ... S_{deg(C_n)}$$
Since if the two subgroups are subgroups of each other, and finite, then they must be equal.
I proceed with the first phase trying to show that
$$ Gal\left( \frac{F}{\Bbb{Q}} \right) \le S_{deg(C_1)} \times S_{deg(C_2)} ... S_{deg(C_n)}$$
Suppose we have roots $r_1, r_2$ such that $r_1$ is a solution to $C_1 = 0$ and the other solve $C_2 = 0$. And an automorphism $\phi | \phi(r_1) = r_2$ which is some permutation not contained in the aforementioned set.
Here's where I start to get stuck. I have intuition that if M is a splitting field for $C_1$ then somehow the existence of such a $\phi$ tells us that once we extend $Q$ to $M$ then we suddenly can solve $C_2$ over the same field (which I want to say is impossible but a counter example is given below: ) $$ C_1 = ((x^2 - 1)^2 -2) , C_2 = x^2 - 2$$
@Gerry Myerson mentions in the comments that if such $\phi$ exists that maps $r_1 \rightarrow r_2$ then it must be the case that $C_1 = C_2$ as both roots must have the same minimal polynomial. So we conclude that this cannot happen. (I would like to find a proof of this fact though), which is given below:
Galois group permutation of roots
Thus we conclude the the group structure of the automorphisms is a subgroup of the permutations of the irreducible polynomials
So now going the other way, we wish to show:
$$ Gal\left( \frac{F}{\Bbb{Q}} \right) \ge S_{deg(C_1)} \times S_{deg(C_2)} ... S_{deg(C_n)}$$
Suppose there was a permutation of the roots contained on the right side that didn't have an associated automorphism. Let this permutation be $pi \in S_{\theta}$. We note that:
It's simply false:
Galois group of an irreducible polynomial
So we only have the Galois group as a subgroup but not the other way.