Let the number of automorphisms of $k=\mathbb Q[x]/(f)$ be $A$ and the distinct subfields in $\mathbb C$ isomorphic to $k$ be $K_1, K_2, \ldots, K_n$. Prove that $$\text{deg}(f) = n \cdot A.$$
One way would be to show that $K_i$ partition the complex roots of $f$ into sets of equal sizes. But I believe that another proof could be contrived using the orbit-stabilizer theorem, but I haven't yet been able to define a proper action that connects the pieces. Is there a way to use group actions to prove the above statement?