Suppose that our polynomial is $x^5-1$, thus the splitting field is $\mathbb{Q}(\gamma)$ where $\gamma$ is a primitive 5-th foot of unity. Then our basis for the extension field will be:
$\{1, \gamma, \gamma^2, \gamma^3, \gamma^4\}$
How would one find a Galois group for this extension? I can see that 1 needs to be fixed, since it is a member of $\mathbb{Q}$, then we can create the maps:
$\sigma_1 = id$
$\sigma_2: \gamma \rightarrow \gamma^2$
$\sigma_3: \gamma \rightarrow \gamma^3$
$\sigma_4: \gamma \rightarrow \gamma^4$
$\sigma_5: \gamma^2 \rightarrow \gamma^3$
$\sigma_6: \gamma^3 \rightarrow \gamma^4$
Does that seem right?
In general for cyclotomic extensions, $K=\Bbb Q(\zeta_n)$, your automorphism is totally determined by what it does on $\zeta$ since that generates the extension. This field is just $\Bbb Q[x]/(\Phi_n(x))$ with $\Phi_n(x)$ the $n^{th}$ cyclotomic polynomial. So an automorphism is just sending one primitive root to another, so there are at most $\varphi(n)$ such automorphisms. However, raising $\zeta_n^k$ for $k\in\Bbb Z/n\Bbb Z^*$ gives $\zeta_n'$ another primitive $n^{th}$ root of $1$, so each $k\in\Bbb Z/n\Bbb Z^*$ gives rise to an automorphism of the group. This induces an injection
$$\Bbb Z/n\Bbb Z^*\to \text{Gal}(K/\Bbb Q)$$
but then since the target has cardinality at most the cardinality of the domain, this must be an isomorphism.
In your case, this exactly reduces to $\gamma\mapsto \gamma^k, \quad 1\le k\le 4$ are all the automorphisms of your field.
Your $\sigma_5$ note has $\sigma_5(\gamma^1)=\sigma_5(\gamma^6)=\sigma_5(\gamma^2)^3=\gamma^{3\cdot 3}=\gamma^4$ so this is the same as $\sigma_4$. And similarly your $\sigma_6$ has $\sigma_6(\gamma^1)=\sigma_6(\gamma^6)=\sigma_6(\gamma^3)^2=\gamma^{2\cdot 4}=\gamma^3$ so it's the same as $\sigma_3$.