Suppose $E=F(S)$.
Since $E$ is a field, it must contain the sum, product and difference of any two elements of $F\cup S$. Furthermore it must contain the sum, product and difference of any of the resulting expressions, etc.
It is clear that since this process of constucting elements using $+,-,\bullet$ started at $F\cup S$, and every element created in this process can be created in a finite number of steps, we have created every element in $F[S]$, the polynomial ring over $F$ in the commuting indeterminates $x \in S$.
But being a field, $E\hspace{1mm}$ must also contain the quotient of any two such expressions (so that the denominator is nonzero). The resulting structure, also denoted $F(S)$, is the field of rational functions/ expressions over $F$ in the commuting indeterminates $x \in S$. Since this is a field, we have by the minimality of $E$, that $E$ equals this field, that is:
$$E=F(S)[\text{field generated by $F$ and $S$}]=F(S)\text{[field of rational functions]}$$
From this, it is clear that any element $\alpha \in E$ is constructed from only a finite number of elements of $S$, that is $\alpha \in E(S')$ for some finite $S' \subseteq S.$
1)) As noted in the comments, the complex conjugation $\sigma$ is a non-identity field automorphism of $\Bbb C$. We have $\sigma(\alpha)=\alpha\in K$, $\sigma(\xi)=1/xi\in K$, so $\sigma(K)\subset K$. Then $K=\sigma(\sigma(K))\subset \sigma(K)$, so $K=\sigma(K)$.
2)) Let $\sigma$ be any field automorphism of $\Bbb C$. Then $\sigma(1)=1$. Let $n$ be any natural number. Since $n=1+1+\dots+1$ (the sum of $n$ $1$’s), $\sigma(n)= \sigma(1)+ \sigma(1)+\dots+\sigma(1)= 1+1+\dots+1=n$. Since $0=\sigma(0)=\sigma(n-n)= \sigma(n)+ \sigma(-n)$, we have $\sigma(-n)=-\sigma(n)=-n$. Let $m$ be any natural number. Then $\sigma(n/m)\sigma(m)=\sigma(n)$, so $\sigma(n/m)= \sigma(n)/ \sigma(m)=n/m$. That is $\sigma(q)=q$ for each $q\in\Bbb Q$.
Since $\xi\in K$ and $K$ is a field, $K$ contains all powers of $\xi$, which are all complex roots of an equation $x^5=1$. Since
$1=\sigma(1)=\sigma(\xi^5)= \sigma(\xi)^5$, $\sigma(\xi)$ is such root, so $\sigma(\xi)\in K$. Since $\alpha\in K$ and
all complex roots of the equation $x^5=1$ are contained in $K$, $K$ contains all complex roots of the equation $x^5=\alpha^5=2$.
Since $2=\sigma(2)=\sigma(\alpha^5)= \sigma(\alpha)^5$, $\sigma(\alpha)$ is such root, so $\sigma(\xi)\in K$.
Since $K=\Bbb Q[\alpha,\xi]$, and $\sigma(\Bbb Q)\subset \Bbb Q\subset K$, $\sigma(\alpha)\in K$, and $\sigma(\xi)\in K$, we have $\sigma(K)\subset K$. Then $K=\sigma(\sigma(K))\subset \sigma(K)$, so $K=\sigma(K)$.
3)) Since $K=\Bbb Q[\alpha,\xi]$ is an extension of $\Bbb Q$ by finitely many elements, algebraic over $\Bbb Q$, $K$ is a finite extension of $\Bbb Q$. Thus, by the answer to the previous question, it suffices to put $E=K$.
4)) If $\sigma$ is complex conjugation then $\sigma(\alpha\xi)=\alpha/\xi\ne\alpha\xi$.
Best Answer
I am sure that $S=5$, it is the most likely answer to that conundrum.
Note that $(\alpha e)^5=2$, so the minimal polynomial of this particular element is $x^5 - 2$ (shown irreducible by Eisenstein on the shifted polynomial) i.e. the extension $\mathbb Q(\alpha e)$ has degree $5$.
Furthermore, the other roots of $x^5-2$ are $\alpha,\alpha e^2,\alpha e^3$ and $\alpha e^4$.
The key ideas used in the proofs of all the above questions is : every automorphism sends a root of a polynomial, to another root of the same polynomial.
1
Suppose that $\sigma(K) = K$ and $\sigma \neq id$. Since $\sigma(\mathbb Q) = \mathbb Q$, we cannot have $\sigma(\alpha e) = \alpha e$ since the set fixed by $\sigma$ would then be the whole of $K$ , since it is a field containing $\mathbb Q$ and $\alpha e$.
Therefore, $\sigma(\alpha e) \neq \alpha e$, but it must be a root of $x^5-2=0$, hence is one of $e,\alpha e^2,\alpha e^3,\alpha e^4$. We show that none of these can belong in $K$.
Indeed, suppose that $\alpha e^n$ belongs in $K$ for $n \not \equiv 1 \mod 5$. Then, we know that $\alpha e \in K$, so taking the quotient, $e^l \in K$ for some $l$ not a multiple of $5$. By Bezout's lemma there is a $d$ with $dl \equiv 1 \mod 5$, so $(e^{l})^d = e^{ld} = e$ is also in $K$.
But then, $K$ is an extension of $\mathbb Q(e)$ so $[K:\mathbb Q]$ must be a multiple of $[\mathbb Q(e) : \mathbb Q]$. The former is $5$, and the latter is $4$, by the minimal polynomial of $e$ being $x^4+x^3+x^2+x+1$, so this is not true.
Consequently, no other root of $x^5-2$ belongs in $K$, therefore $\sigma(K) \neq K$ for any $\sigma \neq id$.
2
This one is pretty clear : any field homomorphism from $K$ is defined by what it does with $\alpha e$ because it always fixes the rational numbers. So simply define $\sigma : K \to \mathbb Q(\alpha)$ by $\alpha e \to \alpha$. This is well defined and extends to $\mathbb C$, via fixing all other elements of $\mathbb C$, and this qualifies as an automorphism of $\mathbb C$ which sends $K$ to a field not equal to $K$.
3
The image of $\sigma : K \to \mathbb C$ must contain a root of $x^5-2 = 0$. If the image equals $K$ then this root can only be $\alpha e$ as shown in part $1$. Consequently $\sigma(\alpha e) = \alpha e$.
4
This one is easy : consider the field consisting of all roots of $x^5-2 = 0$ i.e. $K(\alpha,\alpha e,\alpha e^2,\alpha e^3,\alpha e^4)$. Any automorphism must send $\alpha e$ to a root of $x^5-2 = 0$ but this extension contains all the roots, so it will contain the image of $K$ under any automorphism of $\mathbb C$.