Explicit automorphisms of the splitting field of a certain quartic over the rationals. So I'm working with $\alpha = \sqrt{5+\sqrt{5}}$ and $E=\mathbb{Q}(\alpha)$. The minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $f(x) = x^4 -10x^2 +20$ and I've determined that $E$ is its splitting field. Let $G={\rm Gal}(E/\mathbb{Q})$. Now, $E$ is a finite separable extension, and it's also normal since it's the splitting field of $f$, so we know then that $[E:\mathbb{Q}] = 4 = \left|G\right|$. So either $G\cong \mathbb{Z}_4$ or $G\cong\mathbb{Z}_2\times\mathbb{Z}_2$. Finding the actual automorphisms in $G$ is my problem. I've looked at this  question and the last answer given by oliverjones, but I don't understand how he determines the automorphisms like that. I would appreciate any help, either explaining the answer of oliverjones or a standalone answer/hint to my problem.
 A: As in the linked question, it is not hard to determine that the roots of $f$ are
$$\alpha_1:=\sqrt{5+\sqrt{5}},\qquad\alpha_2:=\sqrt{5-\sqrt{5}},\qquad\alpha_3:=-\sqrt{5+\sqrt{5}},\qquad\alpha_4:=-\sqrt{5-\sqrt{5}}.$$
You noted that $|\operatorname{Gal}(E/\Bbb{Q})|=4$ so there are $4$ field automorphisms of $E$, and they are determined by permutations of these roots. We have the identities
$$\alpha_1+\alpha_3=\alpha_2+\alpha_4=0\qquad\alpha_1\alpha_2=\alpha_3\alpha_4=2\sqrt{5},\qquad\alpha_1\alpha_4=\alpha_2\alpha_3=-2\sqrt{5},$$
and it is easily checked that $\alpha_1\alpha_3\neq\pm2\sqrt{5}$ and $\alpha_2\alpha_4\neq\pm2\sqrt{5}$, and similarly
$$\alpha_1+\alpha_2\neq0,\qquad\alpha_1+\alpha_4\neq0,\qquad\alpha_2+\alpha_3\neq0,\qquad\alpha_3+\alpha_4\neq0.$$
So then the first chain of equalities tells us that $\sigma(\alpha_1)+\sigma(\alpha_3)=0$, and hence
$$\sigma(\alpha_1)=\alpha_1\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_3,$$
$$\sigma(\alpha_1)=\alpha_2\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_4,$$
$$\sigma(\alpha_1)=\alpha_3\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_1,$$
$$\sigma(\alpha_1)=\alpha_4\qquad\Rightarrow\qquad\sigma(\alpha_3)=\alpha_2.$$
In the first and third cases we see that $\alpha_1\alpha_3$ is fixed by $\sigma$, meaning that
$$\sigma(-5-\sqrt{5})=\sigma(\alpha_1\alpha_3)=\alpha_1\alpha_3=-5-\sqrt{5},$$
from which it follows that $\sigma(\sqrt{5})=\sqrt{5}$. So by the second chain of equalities above we get
$$\sigma(\alpha_2)=\sigma\left(\frac{2\sqrt{5}}{\alpha_1}\right)=\frac{2\sqrt{5}}{\sigma(\alpha_1)}\qquad\text{ and }\qquad\sigma(\alpha_4)=\sigma\left(-\frac{2\sqrt{5}}{\alpha_1}\right)=-\frac{2\sqrt{5}}{\sigma(\alpha_1)},$$
which shows that $\sigma$ is determined entirely by where it sends $\alpha_1$. Similarly, in the second and fourth cases we see that $\sigma(\alpha_1\alpha_3)=\alpha_2\alpha_4$, hence that
$$\sigma(-5-\sqrt{5})=\sigma(\alpha_1\alpha_3)=\alpha_2\alpha_4=-5+\sqrt{5},$$
from which it follows that $\sigma(\sqrt{5})=-\sqrt{5}$. So just as before we get
$$\sigma(\alpha_2)=\sigma\left(\frac{2\sqrt{5}}{\alpha_1}\right)=-\frac{2\sqrt{5}}{\sigma(\alpha_1)}\qquad\text{ and }\qquad\sigma(\alpha_4)=\sigma\left(-\frac{2\sqrt{5}}{\alpha_1}\right)=\frac{2\sqrt{5}}{\sigma(\alpha_1)},$$
which again shows that $\sigma$ is determined entirely by where it sends $\alpha_1$.
Hence these are the four automorphisms of $E$ over $\Bbb{Q}$. Now it remains to figure out what $\sigma(\alpha_2)$ and $\sigma(\alpha_4)$ are in each of these cases, but that is a matter of some handy computation.

Most likely you will soon learn some more advanced methods for determining the structure of the Galois group of a field extension. But looking at symmetric expressions in the roots of a generating polynomial that take values in the base field (or intermediate extensions) is a very concrete approach that also makes very clear what the automorphisms are, in stead of only determining the Galois group up to isomorphism. So if I may I'd recommend trying this approach on another extension for yourself.
A: The roots of this polynomial are $\sqrt{5+\sqrt{5}}$,  $-\sqrt{5+\sqrt{5}}$, $\sqrt{5-\sqrt{5}}$ and $-\sqrt{5-\sqrt{5}}$  the discriminant of the polynomial is not squart so th galois group is cyclic.
other method is to look at how many are there intermediate extension:
the intermediate extensions are given by the coefficients of the
following polynomial
$(x-\sqrt{5+\sqrt{5}})(x+\sqrt{5+\sqrt{5}})=x^2-5-\sqrt{5}$
$(x-\sqrt{5+\sqrt{5}})(x-\sqrt{5-\sqrt{5}})= x^2-x\sqrt{\left(
5-\sqrt{5}\right) }-\sqrt{\left( 5+\sqrt{5}\right) }x+\sqrt{\left(
5+\sqrt{5}\right) }\sqrt{\left( 5-\sqrt{5}\right) }$
$(x-\sqrt{5+\sqrt{5}})(x-\sqrt{5-\sqrt{5}})=x^2-x\sqrt{\left( 5-\sqrt{5}%
\right) }-\sqrt{\left( 5+\sqrt{5}\right) }x+\sqrt{\left( 5+\sqrt{5}\right) }%
\sqrt{\left( 5-\sqrt{5}\right) }$.
The first polynomial give $\Bbb{Q}(\sqrt{5})$
the second  $\Bbb{Q}(\sqrt{( 5-\sqrt{5}) }-\sqrt{( 5+\sqrt{5}) },\sqrt{(5+\sqrt{5}) }\sqrt{( 5-\sqrt{5}) })$ 
then contain $\sqrt{( 5-\sqrt{5}) }+\sqrt{( 5+\sqrt{5}) }$ and so contian $\sqrt{(
5+\sqrt{5}) }$ the field  is the same as  the whole field $\Bbb{Q}(\alpha)$
for the third polynomial, the field is also the whole
field $\Bbb{Q}(\alpha)$, and so there are only an field intemidiate
and the galois group for the whole field is cyclique $Z_4$
