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.

Yes, of course there is, and this is one of the main results in fields extensions-Galois Theory. 
Theorem: If $\;K/F\;$ is a fields extension of degree $\;n\;$ and $\;S,\,F\subset S\subset K\;$  is the separable closure of this extension, then there are $\;[S:F]\;$different embeddings $\;K\mapsto\overline{\Bbb F}\;$ , when this last is any algebraically closed field containing $\;F\;$ ( we can take the algebraic closure of $\;F\;$) .
If we assume the extension $\;K/F\;$ is finite and separable, then the number of embeddings equals $\;n=[K:F]\;$ iff $\;K/F\;$ is normal, which would mean the extension $\;K/F\;$ is Galois. In this last case, any such embedding is an automorphism of $\;K/F\;$ , meaning $\;\sigma K=K\;$ for any embedding $\;\sigma:K\to\overline F\;$, and $\;\sigma f=f\;,\;\;\forall\,f\in F\;$ .
