Is there any relation between the number of distinct embeddings of a number field into $\mathbb C$ and the number of $\mathbb Q$ automorphisms of it? Suppose that $K$ is a finite extension of $\mathbb Q$, say of degree $n$. By the primitive element theorem, $K=\mathbb Q(\alpha)$. Then $\alpha$ has $n$ conjugates and we correspondingly get $n$ embeddings of $K$ into $\mathbb C$. But I believe that all these embeddings need to be have the same images in $\mathbb C$ (as an example, $\mathbb Q(\sqrt 2)$ and $\mathbb Q(-\sqrt 2)$ are really the same field). Is there any relation between the number of embeddings with distinct images and the order of the group $Aut(K/\mathbb Q)$? On working a few examples, it seems that if $|Aut(K/\mathbb Q)|=a$ and there are $b$ embeddings with distinct images in $\mathbb C$, then $ab=n$, the degree of the extension. Is this true? If so, how can I prove it? I don't seem to be having much success. I'd appreciate some help.

For an algebraic extension $K/\mathbb{Q}$ of degree $n$ there are $n$ embeddings of $K$ into $\mathbb{C}$, see 
(The number of) embeddings of an algebraic extension of $\mathbb{Q}$ into $\mathbb{C}$. For example, let $K=\mathbb{Q}(\sqrt[4]{2})$. Then $a=|Aut_{\mathbb{Q}}(K)|=2$, $b=4$ and $n=4$. Hence $ab\neq n$.
