Suppose that we have Galois covering $$X\longrightarrow P^{1}\_{\mathbb{Q}}$$ defined over the rationals which is cyclic, in the sense that the Galois group associated to the covering is a cyclic group. 
Then $X$ can not be $P^{1}\_\mathbb{Q}$. I read this result somewhere. Is this result true? I tried this over the complex numbers and I found that over complex numbers it is true:


$P^{1}\_{\mathbb{C}} \longrightarrow P^{1}\_{\mathbb{C}}$ defined by $z$ going to $z^{n}$. 



Question related to Galois covering of Projective line over rational numbers
*Your (!)* [MO-question](https://mathoverflow.net/questions/244541/action-of-mathbbz-3-mathbbz-on-p1) (together with the answer/comments) shows that the group of order $3$ is also a Galois group of a Galois cover $\mathbb P^1\_{\mathbb Q}\to\mathbb P^1\_{\mathbb Q}$. Besides this and order $2$, the only other possible order is $4$ which can be seen as follows:


As $\text{Aut}\_{\mathbb Q}(P^1\_{\mathbb Q})=\text{PGL}\_2(\mathbb Q)$, your question reduces to finding elements of finite order in $\text{PGL}\_2(\mathbb Q)$. Let $A\in\text{GL}\_2(\mathbb Q)$ represent an element of order $m$ in $\text{PGL}\_2(\mathbb Q)$. Then $A$ is diagonalizable over a quadratic extension of $\mathbb Q$. Let $a$ and $b$ be the eigenvalues of $A$. Then $\zeta=a/b$ is a primitive $m$-th root of unity. Together with the fact that $\zeta$ has degree at most $2$ over the rationals, one gets $m\le4$ (e.g. by using the irreducibility of the cyclotomic polynomials, or by more elementary means).


The order $4$ indeed arises, as $z\mapsto(z-1)/(z+1)$ has order $4$ on $\mathbb P^1\_{\mathbb Q}$.


