Suppose $G$ acts on $\Omega.$ It's easy to see that if $g \in G$ is arbitrary, then the function $\sigma_g: \Omega \to \Omega$ defined by $(\alpha)\sigma_g = \alpha \cdot g$ has an inverse: the function $\sigma_{g^{-1}}$. Therefore, $\sigma_g$ is a permutation of the set $\Omega$, which means $\sigma_g$ is both injective and surjective, and thus $\sigma_g$ lies on the symmetric group $\text{Sym}(\Omega)$ consisting of all permutations of of $\Omega$. [...]
My question is what's the function $\sigma_{g^{-1}}$? Is it the function $\sigma_{g^{-1}}: \Omega \to \Omega$ defined by $$(\alpha)\sigma_{g^{-1}} = \alpha \cdot g^{-1}$$
In which case we would have $\alpha \sigma_g \alpha \sigma_{g^{-1}} = \alpha g \alpha g^{-1}$ which doesn't seem to be the identity. I think I'm being confused by the notation.
The only way this would make sense to me is that if I define $\sigma_{g^{-1}}: \Omega \to \Omega$ by $$\sigma_{g^{-1}} = (\alpha \cdot g)^{-1} = g^{-1}\alpha^{-1}.$$
But that does not appear to be consistent with the author's notation.
Note that $\sigma$ is a function. So it only makes sense for function composition. Then $(\alpha)(\sigma_g \circ \sigma_{g^{-1}}) = (\alpha \cdot g)\cdot g^{-1} = \alpha$.