Well, let's see. I hope I don't have to prove that $\mathbb Q$ has only one endomorphism. I skip that. Well, given that, it is obvius that $\mathbb Q\subset K$ is an strict inclussion. One of the endomorphism has to be the identity $id$ (obviusly), and the other let's call it $\phi$, then $\phi\circ\phi=id$. Take $\alpha\in K\setminus\mathbb Q$, then if $\beta=\phi(\alpha)$, then $\phi(\beta)=\alpha$. Note that $(x-\alpha)(x-\beta)\in\mathbb Q[x]$, because $\alpha+\beta=\phi(\alpha+\beta)$ and $\alpha\beta=\phi(\alpha\beta)$. Then, $\alpha=\frac{-b\pm c\sqrt{d}}{2a}$ for some integers $a,b,c,d$, with $d$ square free (otherwise you just take the square out of the root and multiply it to c). But then, $K=\mathbb Q(\alpha)=\mathbb Q(-b+c\sqrt{d})=\mathbb Q(c\sqrt{d})=\mathbb Q(\sqrt{d})$, because $2a,-b,c$ are rationals!