= Solution
Let $f$ be the common minimal <polynomial>. Evaluation gives
$$
\mathbb Q[X]/(f)\cong\mathbb Q(\alpha),
\qquad
\mathbb Q[X]/(f)\cong\mathbb Q(\beta),
$$
so composing the two isomorphisms gives $\mathbb Q(\alpha)\cong\mathbb Q(\beta)$ and sends $\alpha$ to $\beta$.
Take $\alpha=\beta=\sqrt[4]{2}$. The field is real and contains the two roots $\pm\alpha$ of $X^4-2$, but not $\pm i\alpha$. Hence exactly two $\mathbb Q$-automorphisms are possible, while the degree is four.
In general, after choosing one isomorphism $\phi:\mathbb Q(\alpha)\to\mathbb Q(\beta)$, every other one is uniquely $\phi\circ\sigma$ for a $\mathbb Q$-automorphism $\sigma$ of $\mathbb Q(\alpha)$. Thus
$$
k=|\operatorname{Aut}_{\mathbb Q}(\mathbb Q(\alpha))|.
$$
By the <automorphism-count divisibility theorem>, this order divides $[\mathbb Q(\alpha):\mathbb Q]$. Therefore $k$ always divides the degree.
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