There is a slight divergence of nomenclature. Everyone agrees on what $\mathrm{Aut}(E/F)$ is. The question is what to call it.
Some books (e.g., Hungerford, Rotman's Galois Theory), always refer to $\mathrm{Aut}(E/F)$ as the "Galois group" of $E$ over $F$ (or of the extension), whether or not the extension is a Galois extension.
Other books (e.g., Lang), use the generic term "automorphism group" to refer to $\mathrm{Aut}(E/F)$ in the general case, and reserve the term Galois group exclusively for the situation in which $E$ is a Galois extension of $F$.
So, in Lang, even just saying "Galois group" already implies that the extension must be a Galois extension, that is, normal and separable. In Hungerford, just saying "Galois group" does not imply anything beyond the fact that we are looking at the automorphism of the extension.
Wikipedia is following Convention 2; your book is following convention 1.
There is also the question of whether to admit infinite extensions or not. A lot of introductory books only consider only finite extensions when dealing with Galois Theory, and define an extension to be Galois if and only if $|\mathrm{Aut}(E/F)| = [E:F]$. This definition does not extend to the infinite extension, so the definitions are restricted to finite (algebraic) extensions, with infinite extensions not considered at all. Other characterizations of an extension being Galois (e.g., normal and separable) generalize naturally to infinite extensions, so no restriction is placed. Likewise, some books explicitly restrict to algebraic extensions, others do not; but note that most define "normal" to require algebraicity, because it is defined in terms of embeddings into the algebraic closure of the base field, so even if you don't explicitly require the extension to be algebraic in order to be Galois, in reality this restriction is (almost) always in place.
This is not such a big deal as it might appear, because one can show that an arbitrary (possibly infinite) Galois extension $E/F$ is completely characterized in a very precise sense by the finite Galois extensions $K/F$ with $F\subseteq K\subset E$ with $[K:F]\lt\infty$, as the automorphism group $\mathrm{Aut}(E/F)$ is the inverse limit of the corresponding finite automorphism groups.
It cannot be $\mathbb{Z}_2 \times \mathbb{Z}_2$ for the following reason:
The splitting field for $f(x) = x^4-5$ is $K = \mathbb{Q}[\sqrt[4]{5}, i]$. We have the tower of fields $\mathbb{Q} \subset \mathbb{Q}[\sqrt[4]{5}] \subset K$. We have $[\mathbb{Q}[\sqrt[4]{5}]:\mathbb{Q}] = 4$, and since $i \notin \mathbb{Q}[\sqrt[4]{5}]$, then $x^2+1$ is irreducible in this field, so we must have $[K:\mathbb{Q}[\sqrt[4]{5}]] = 2$. (See footnote)
Therefore, $[K:\mathbb{Q}] = [K:\mathbb{Q}[\sqrt[4]{5}]] \cdot [\mathbb{Q}[\sqrt[4]{5}]:\mathbb{Q}] = 8$. Recall that the order of $\operatorname{Gal}(f/\mathbb{Q})$ is equal to the degree of the splitting field of $f$ over $\mathbb{Q}$, so $|\operatorname{Gal}(f/\mathbb{Q})| = 8$. Further, since $f$ has $4$ roots, we have $\operatorname{Gal}(f/\mathbb{Q}) \subset S_4$. Note that subgroups of order $8$ inside of $S_4$ are Sylow $2$-subgroups, and all Sylow $p$-subgroups are isomorphic to each other for a given $p$, so it will suffice to simply find a single subgroup of order $8$ inside of $S_4$ and classify it.
Alternatively, you can list the automorphisms by hand, and you'll know your list is exhaustive once you've written down $8$ of them, then compare to groups you're already familiar with (e.g. $\mathbb{Z}_8, \ \mathbb{Z}_4 \times \mathbb{Z}_2, \ D_4, \ \mathcal{Q}_8$, etc).
Footnote: The degree of an extension $F[\alpha]$ over $F$ is equal to the degree of an irreducible polynomial in $F[x]$ with $\alpha$ as a root. So for example, $\mathbb{Q}[\sqrt[4]{5}]$ is a degree $4$ extension over $\mathbb{Q}$ since $\sqrt[4]{5}$ is a root of $g(x) = x^4-5$, which is an irreducible polynomial.
Best Answer
Without deep machinery, any field automorphism $\phi$ of $\Bbb C$ that leaves $\Bbb R$ pointwise fix is determined by $\phi(i)$ as we must have $\phi(x+iy)=x+\phi(i)y$. Also $$0=\phi(0)=\phi(1+i^2)=1+\phi(i)^2$$ implies that $\phi(i)\in\{i,-i\}$. Hence indeed there are exactly two such automorphisms: the identity and complex conjugation. So $G(\Bbb C/\Bbb R)$ is the group of order $2$.