Is it possible that each element of Galois group preserve a root? Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial. Denote by $K$ the splitting field for $f(x)$. Let $G$ be Galois group of $K$ over $\mathbb{Q}$. Let $\alpha_1, \dots, \alpha_n \in K$ be all the roots of $f(x) = \Pi (x - \alpha_i)$.
Question: Does there exist such an $f(x)$ that for each $g \in G$ there exist $\alpha_i$ such that $g( \alpha_i ) = \alpha_i$?
Edited: I am interested in the case $\deg f \geq 2$.
 A: No, unless $G$ is trivial.
The action of the Galois group of an irreducible polynomial is transitive and every transitive permutation group on a finite set has an element that  fixes no point.
A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$No when $G \ne \{ 1 \}$, that is, when the degree of $f$ is greater than $1$, a requirement just added by OP.
The reason is that the Galois group acts transitively on the set $\Omega$ of the roots, as $f$ is irreducible.
Thus all stabilizers are conjugate. And it easy to prove that a finite group cannot be the set-theoretic union of the conjugates of a proper subgroup - I have appended the standard proof below.

In fact, suppose $G$ is a finite group, $H < G$, and consider the set
$$
\bigcup_{g \in G} H^{g},
$$
where $H^{g} = g^{-1} H g$. We have
$$
\Size{\bigcup_{g \in G} H^{g}} 
= 1 +
\Size{\bigcup_{g \in G} (H^{\#})^{g}}
\le
1 + \Size{H^{\#}} \cdot \Size{G : N_{G}(H)}
\le\\\le
1 + (\Size{H} - 1) \cdot \Size{G : H}
=
\Size{G} - \Size{G : H} + 1 
< \Size{G},
$$
where I have used the fact that $H < G$, so that $\Size{G : H} > 1$.
Here $H^{\#} = H \setminus \Set{1}$.
