Show: If some $x_i \neq x_1$ occurs as $\sigma(x_1)$ for a $\sigma \in$ Gal$ (E:F)$ then each $x_i\neq x_1$ occurs as $\sigma(x_1)$ I'm currently studying Galois Theory (or am trying to do so), and since almost two weeks I try over and over again to solve a particular question from a textbook (J. Stillwell, Elements of Algebra, p. 132).
Task: Consider the extension $E=F(x_1)$ of $ F=\mathbb{Q}(a_0,...,a_4)$. $x_1$ is a root of the quintic $x^5+a_4 x^4 + ... + a_0 =0$ (the other four roots being depicted as $x_2$ to $x_5$). Show
(i) $\sigma \in$ Gal$(E:F)$ is determined by the value $\sigma(x_1)$.
(ii) $\sigma(x_1)$ is a root of $x^5+a_4 x^4 + ... + a_0 =0$, hence $|$Gal$(E:F)| <= 5$.
(iii) If some $x_i \neq x_1$ occurs as $\sigma(x_1)$ for a $\sigma \in $ Gal$ (E:F)$ then each $x_i\neq x_1$ occurs as $\sigma(x_1)$ 
It turned out that irreducibility of the quintic over F, although not stated explicitly,
is absolutely necessary for claim (iii) to be true, see comments below.

I had little trouble to show (i) and (ii), but (iii) is giving me a headache permanently. Here is what I tried:
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(Trial A) 
Suppose $\sigma(x_1) = x_2$, and $\sigma \in $ Gal$(E:F)$. Then:  
$\Rightarrow \sigma(f) = f $ for all $ f\in F$ by definition of Galois Groups, and $x_i \in E$ since $\sigma$ is an automorphism on E.
$\Rightarrow $ There must be a rational function $q$ of $x_2$ with $x_1 = q(x_2)$, because $x_1$ must be the result for one $\sigma(q(x_1)) = q(\sigma(x_1)) = q(x_2)$
$\Rightarrow x_2 = \sigma(x_1)=\sigma(q(x_2)) = q(x_1)$ and thus: $\sigma^2(x_1)=\sigma(x_2)=q(\sigma(x_1))=q(x_2)=x_1$. Thus $\sigma$ is its own inverse.
This was not correct, see comment below. 
Thus the next conclusion is wrong, also.

$\Rightarrow $ Now $\{1,\sigma\}$ is closed under multiplication, it is easy to see that it is a group. Although I don't know whether it actually *is* the Galois-Group of E over F, it could be. At least I don't see a reason from here why it shouldn't be fine, or why I would have to include more $x_i$ necessarily.
EDIT#2: From this point I tried to construct a counterexample to the statement to be shown, in order to see which are the obstacles. If I have two roots $x_1$ and $x_2$ with a corresponding automorphism, I can choose the other three roots as definitely lying outside of $\mathbb{Q}(a_0,...,a_4,x_1)$. Then with
$ (x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)$ I have a polynomial with three roots outside of $E$, contradicting the statement to be shown. Reason: if $x_i \notin  E$ there cannot be an automorphism $\sigma \in $ Gal$ (E:F)$ with $\sigma(e) = x_i$ for some $e \in E$. However, the field $F=\mathbb{Q}(a_0, ...a_4)$ is determined by the choice of $x_3, x_4$ and $x_5$, via the $a_i$. Thus it is still possible that I cannot find roots outside $F$ just by this feedback behavior. This would again point towards some argument of the kind: Because two roots are fixed, the others will be in $\mathbb{Q}(a_0, ...a_4)$ since they can be calculated by $x_1$, $x_2$ and the $a_i$. What to do from here...?
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(Trial B)
The next try was to go over the symmetric elementary polynomials $a_1,...,a_4$. If I assumed two roots $x_1$ and $x_2$ to lie in $E$, is it possible to calculate the other roots from the $a_i$? Well, it seems to be. That means that with $x_1$ and $x_2$ in $E$ all the others would lie in $E$, too, and it remains to show that the corresponding automorphisms exist. But I only could hack it with mathematica, and the result is rather complicated. I doubt that this was the authors intention. Additionally, I cannot see why then it shouldn't be possible to calculate four roots out of $x_1$, with the five elementary polynomials.
EDIT#1: I discovered a serious misinterpretation in my Trial B. I can compute expressions $f(a_0, ...,a_4)$ for each $x_3, x_4$ and $x_5$, respectively. But they contain $\sqrt{}$ and $\sqrt[3]{}$ operations, and thus are not in $E = \mathbb{Q}(a_0,...,a_4,x_1)$. If I'm right, this rules out Trial B as an appropriate way to show what is asked. It must have something to do with the automorphisms in general, some symmetry-argument maybe, but I don't know.
As I see it according to the solution of Jyrki, the a_i do indeed
determine which x_i are in E and what the automorphisms are. 
But this is shown via the general properties of E, not by direct calculation.

Please, can someone help me?
 A: Recapping the setting: $F=\Bbb{Q}(a_0,a_1,a_2,a_3,a_4)$ so the polynomial
$$
p(x)=x^5+a_4x^4+a3x^3+a_2x^2+a_1x+a_0\in F[x].$$
We further denote the zeros of $p(x)$ (in some extension of $F$) by $x_1,x_2,\ldots,x_5$, and also denote $E=F(x_1)$.
Assumption. There exists a non-trivial $F$-automorphism $\sigma$ of $E$.
Extra assumption. The polynomial $p(x)$ is irreducible in $F[x]$.
Claim. Given the extra assumption all the zeros $x_i, i>1,$ are in the field $E$. Furthermore, $\sigma$ is necessarily of order five and it permutes the five roots in a 5-cycle. In particular, given $x_i, i\neq1$, there exists an exponent $\ell$ (depending on $i$) such that $\sigma^\ell(x_1)=x_i$.
Proof. The extra assumption implies that $[E:F]=\deg p(x)=5$. I will heavily use the fact that five is a prime. Most notably, it follows that there are no intermediate fields $L$ such that $F\subset L\subset E$. This is immediate from the multiplicativity of the extension degree in a tower of field extension.
Consider the set $M$ of fixed points of $\sigma$
$$
M=\{z\in E\mid \sigma(z)=z\}.
$$
It is known that $M$ is field (do this as an exercise if you haven't seen it yet - all you need is that $\sigma$ is an automorphism of fields). Because $\sigma$ is not the identity mapping, $M$ is a proper subfield of $E$. OTOH $\sigma$ is an $F$-automorphism of $E$, so $F\subseteq M$. Lack of intermediate fields implies that $M=F$.
Another standard fact is that if $z\in E$ is a zero of $p(x)$ then so is $\sigma(z)$. Consequently all the numbers $x_1,\sigma(x_1),\sigma^2(x_1),\ldots$ are elements of $E$, and they are all zeros of $p(x)$. Because $p(x)$ has at most five zeros in $E$ there exists an exponent $m, 1<m\le5,$ such that $\sigma^m(x_1)=x_1$. (Proof: there must be repetitions among those iterates, say $\sigma^i(x_1)=\sigma^j(x_1)$ for some $i<j$, because $\sigma$ is a bijection we can cancel some, and reduce to the case $i=0$.)
Consider the polynomial
$$
m(x)=\prod_{i=0}^{m-1}(x-\sigma^i(x_1))\in E[x].
$$
Because the sets $\{\sigma^i(x_1)\mid 0\le i<m\}$ and
$\{\sigma^i(x_1)\mid 0<i\le m\}$ are equal. It follows that if we apply $\sigma$ to all the coefficients of $m(x)$ we must get $m(x)$ back. In other words $m(x)\in M[x]=F[x]$. On the other hand, $m(x)$ is a factor of $p(x)$ (all zeros of $m(x)$ are simple and also zeros of $p(x)$). The extra assumption about the irreducibility of $p(x)$ then implies that $m(x)=p(x)$.
In particular $m=5$. Consequently $p(x)$ has five zeros in $E$. The other claims follow from this.
What happens if we drop the extra assumption?
Assume instead that $p(x)=r(x)s(x)$ non-trivially. Without loss of generality
$r(x_1)=0$, and we can assume that $r(x)$ is irreducible over $F$ (i.e. the minimal polynomial of $x_1$ over $F$). If we also assume claim (iii) then all the zeros of $s(x)$ must appear as automorphic images of $x_1$. But those images are also zeros of $r(x)$. In other words, the zeros of $s(x)$ must all be zeros of $r(x)$. It can be shown (we may need a few more pieces of theory here) that all the zeros of $p(x)$ must have the same multiplicity, implying that $\deg r(x)\mid \deg p(x)$. Again, as five is a prime, this leaves us with the possibilities of $r(x)=p(x)$ and the possibility $\deg r(x)=1$. In the latter case $F(x_1)=F$ and we arrive at a contradiction.
