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
Saying "the reals are an extension of the rationals" just means that the reals form a field, which contains the rationals as a subfield. This does not mean that the reals have the form $\mathbb{Q}(\alpha)$ for some $\alpha$; indeed, they do not. You have to adjoin uncountably many elements to the rationals to get the reals.