I'm given the structure $\mathcal{M} = (\mathbb{Q} \setminus \{0\}; =; \cdot)$ and I need to prove that Aut $\mathcal{M}$ is continual. In other words, first of all, I want to find all (or at least most of) bijections $\alpha : \mathbb{Q} \setminus \{0\} \to \mathbb{Q} \setminus \{0\}$ so that $\alpha(a \cdot b) = \alpha(a) \cdot \alpha(b) \;\; \forall a, b \in \mathbb{Q} \setminus \{0\}$.
The problem is, I can't think of any automorphisms except for the identity and $\alpha(x) = \dfrac1x$, of course. I mean, let's take polynomials $\alpha \in \mathbb{Q}[x]$, then only $\alpha(x) = x^n, n \in \mathbb{N}_+$ "saves" multiplication as mentioned above, but it's not a bijection since $\sqrt[n]{y}$ could be irrational. Same for $\alpha(x) = \dfrac1{x^n}$.
For obvious reasons, there's no way logarithms, exponential or trigonometrical functions can be used.
So I'm basically stuck in the beginning of this problem and would be glad if someone helped/gave a clue!