Let $G$ a group acting on $\Omega$. A subset $\Sigma \subseteq \Omega$ is called a base if $$G_{(\Sigma)} := \{ x \in G : \delta^x = \delta \mbox{ for all } \delta \in \Sigma \} = 1$$ (i.e. the pointwise set stabilisator is trivial). This is equivalent to
i) $\Sigma^x$ is a base for each $x \in G$,
ii) for all $x,y \in G$, $\alpha^x = \alpha^y$ for all $\alpha \in \Sigma$ implies $x = y$,
iii) for each $x \ne 1$ there exists some point in $\Sigma$ moved by $x$ (i.e. we have $\alpha^x \ne \alpha$ and $\alpha \in \Sigma$).
Now the question:
Let $G$ be a transitive subgroup of $Sym(\Omega)$ where $\Omega$ is finite, and suppose that $\Sigma$ is a minimal base, show that for each $\alpha \in \Omega$ there are exactly $|\Sigma||G| / |\Omega|$ values of $x \in G$ such that $\alpha \in \Sigma^x$.
Let $\alpha \in \Omega$. As $G$ is transitive and finite we have $|\Omega| = |G : G_{\alpha}|$, hence $$ |G| / |\Omega| = |G_{\alpha}| $$ so we have to show that $|\{ x \in G : \alpha \in \Sigma^x \}| = |\Sigma||G_{\alpha}|$. Now I am trying to establish a bijection between the set $\{ x \in G : \alpha \in \Sigma^x \}$ and $\Sigma\times G_{\alpha}$, but I do not see how to make this map well-defined and being injective and surjective. So any hints? Maybe my attempt is also not very fruitful at all?