Consider $p(x) = (x^5-1)(x^2+1)$. Then, its splitting field is $\mathbb{Q}(e^{\frac{2\pi i}{5}}, i)$.
Thus, $f\in \text{Gal}(\mathbb{Q}(e^{\frac{2\pi i}{5}}, i)/\mathbb{Q})$ maps $\omega = e^{\frac{2\pi i}{5}}$ to any of $\omega^k$ for $k=1,...,4$ and $i$ to $\pm i$.
In that way, I can conclude that $|\text{Gal}(\mathbb{Q}(e^{\frac{2\pi i}{5}}, i)/\mathbb{Q})|= 4 \cdot 2 = 8$.
Now, how do I know which of the groups of order $8$ it is? It might be $\mathbb{Z}_2 \times \mathbb{Z}_4$ because of element orders but I am not sure.
You have found all the elements of the Galois group. This means it wont be too hard to find a structure description of the group by looking at how the elements interact.
An important element is the complex conjugation automorphism. Call it $\tau$. $\tau$ maps $i$ to $-i$ and maps $\omega$ to $\omega^{-1} = \omega^{4}$ Let's consider how $\tau$ interacts with the maps $\sigma_k$ defined by $\sigma_k \omega = \omega^k$, $\sigma_k i = i$.
$\sigma_k \tau (\omega) = \sigma_k (\omega^{-1}) = \omega^{-k}$
$\sigma_k \tau (i) = \sigma_k (-i) = -i$
$\tau \sigma_k (\omega) = \tau (\omega^k) = \omega^{-k}$
$\tau \sigma_k (i) = \tau (i) = -i$
So we have seen that $\tau$ commutes with each $\sigma_k$.
This tells us that $G$ is an inner direct product of $\langle \tau \rangle$ of order 2 with $\langle \sigma_1 \rangle$ of order 4. Thus $$G = C_2 \times C_4.$$