Let $\mathbb{Q} \subset K$ be a finite field extension and $\varphi \in \text{Emb}(K,K)$. I want to show that if $\varphi(x)=y$ then $\varphi (\overline{x}) = \overline{y}$, where $\overline{x}$ denotes complex conjugation. Is the following proof valid?
We know that $\varphi(1)=1$ so $\varphi(n)=n$, $\forall n \in \mathbb{Z}$. Hence $\varphi(q) = q$, $\forall q \in \mathbb{Q}$. Therefore $\varphi$ restricted to $\mathbb{Q}$ must be the identity.
Now $\varphi(x) = \varphi(x_1 + x_2 i) = \varphi(x_1) + \varphi(x_2 i) = x_1 + \varphi(x_2 i)$
So $\varphi(\overline{x}) = \varphi(x_1 - x_2 i) = \varphi(x_1) - \varphi(x_2 i) = x_1 - \varphi(x_2 i)$
Hence we conclude that $\varphi(\overline{x})$ and $\varphi(x)$ are complex conjugates.
Instead of pointing out where the argument breaks, let's try a counter-example. We want an extension where the complex conjugation is not in the center of the Galois group (it does not commute with all the automorphisms).
The simplest example is to take a Galois group isomorphic to the symmetric group $S_3$, which has trivial center, and a classical way to do that is to take the splitting field of $X^3-2$, which is $K=\mathbb{Q}[\sqrt[3]{2},j]$ (where $j$ is a cubic root of unity).
Then there is an automorphism $\varphi$ of $K$ sending $x=\sqrt[3]{2}$ to $y=j\sqrt[3]{2}$, so clearly $\varphi(\bar{x})\neq \bar{y}$.