Galois group $Gal(\mathbb{Q}[\xi]:\mathbb{Q})$

Lets $\xi$ is primitive n-th root of unity over $\mathbb{C}$ and lets $$\mathbb{Q}[\xi]\ni\alpha=\sum_{i=0}^{n-1}a_i{\xi}^i.$$ Consider any element $g\in Gal(\mathbb{Q}[\xi]:\mathbb{Q})$. $$g(\alpha)=g(\sum_{i=0}^{n-1}a_i{\xi}^i)=\sum_{i=0}^{n-1}g(a_i)g({\xi}^i)=\sum_{i=0}^{n-1}g(a_i)(g({\xi}))^i.$$ Question: is it true that $\exists k\in \{0, 1,\dots,n-1 \}:g(\xi) = {\xi}^k$?

Or $g(\xi)=x_0+x_1{\xi}^1+\dots+x_{n-1}{\xi}^{n-1}$ (where $x_0, \dots, x_{n-1}\in \mathbb{Q}$)?

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Yes, it is true that, when $\xi$ is a primitive $n$th root of unity, any $g\in\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})$ must satisfy $g(\xi)=\xi^k$ for some $0\leq k\leq n-1$. In fact, only certain $k$'s can occur: those $k$ which are relatively prime to $n$ (so $k=0$ is out, for example).

Here is a proof. Because $\xi$ is an $n$th root of unity, we have that $\xi^n=1$. Because $g$ is a homomorphism, we have that $$\left(g(\xi)\right)^n=g(\xi^n)=g(1)=1.$$ Thus, $g(\xi)$ is some $n$th root of unity; but the $n$th roots of unity are precisely $\xi^0,\ldots,\xi^{n-1}$, so $g(\xi)$ must be among these $n$ numbers.

If $g(\xi)^r=1$ for some $r<n$, then we can apply $g^{-1}$ to both sides to reach $\xi^r=1$, which contradicts the assumption that $\xi$ is a primitive $n$th root of unity. Thus, $g(\xi)$ cannot be a $r$th root of unity for any $r<n$; in other words, $g(\xi)$ is a primitive $n$th root of unity as well. One can prove that these are the roots of unity $\xi^k$ for which $\gcd(k,n)=1$.

I also wanted to point out that, in your second displayed equation, we can simplify further: each of the $g(a_i)$ is actually equal to $a_i$.

We can represent an arbitrary element of $\mathbb{Q}(\xi)$ as some sum $$\alpha=\sum_{k=0}^{n-1}a_k\xi^k,\qquad a_k\in\mathbb{Q}.$$ Recall that elements $g\in\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})$ are isomorphisms from $\mathbb{Q}(\xi)$ to itself that fix $\mathbb{Q}$, i.e., $g(x)=x$ for any $x\in\mathbb{Q}$. Thus, we actually get $$g(\alpha)=g\left(\sum_{k=0}^{n-1}a_k{\xi}^k\right)=\sum_{k=0}^{n-1}g(a_k)g({\xi}^k)=\sum_{k=0}^{n-1}g(a_k)(g({\xi}))^k=\sum_{k=0}^{n-1}a_k(g({\xi}))^k.$$ This demonstrates a very important point: that an automorphism $g\in\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})$ is determined completely by what it does to $\xi$.

In other words, for any $g_1,g_2\in\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q})$, if it is true that $g_1(\xi)=g_2(\xi)$, then in fact $g_1=g_2$; and in the other direction, once we know what $g$ does to $\xi$, we know what it does to any $\alpha\in\mathbb{Q}(\xi)$.

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