Let $G$ be a transitive permutation group such that $|\mbox{fix}(g)| \in \{0,3\}$ for every nontrivial $g \in G$. Also suppose $|N_G(G_{\alpha}) : G_{\alpha}| = 1$, i.e. $G_{\alpha}$ is the only fixed point of all elements in $G_{\alpha}$ (by this assumption there exists some $g \in G$ such that $G_{\alpha} \ne G_{\alpha}^g$ and $G_{\alpha} \cap G_{\alpha}^g \ne 1$). Also note that by the assumption $|\mbox{fix}(g)| = 2$ for every nontrivial $g \in G_{\alpha}$.
Also assume $\operatorname{gcd}(|\Omega|, |G_{\alpha}|) \in \{1,3\}$ and $G_{\alpha} \cong A_4$ and that $G_{\alpha}$ has three non-regular orbits of sizes $6$, $4$ and $4$. Then it acts on the two orbits of size $4$ as a Frobenius group. Let $V$ be the four-group of $A_4$. As $\alpha$ is the only fixed point of $V$, we have $N_G(V) = V$ and $C_G(V) = V$.
Why do we have $C_G(V) = V$?
As $V$ is normal in $G_{\alpha}$, take some orbit and look at the orbits of $V$, then we see that they must have size $ > 1$ for otherwise $V$ would fix all points of the orbit (the orbits must all have the size size by normality of $V$). As $V$ could have at most two other fixed points but the orbits are larger, this is not possible. Hence $\alpha$ is the only fixed point of $V$. Because if $g \in N_G(V)$, then $g$ permutes the fixed points of $V$ (for if $V \le G_{\beta}$, then $V = V^g \le G_{\beta^g}$, or by direct computation $(\beta^g)^u = \beta^{gu} = \beta^{u'g} = \beta^g$). But as $V$ just has a single fixed point, each $g \in N_G(V)$ must fix $\alpha$, i.e. $N_G(V) \le G_{\alpha}$, and by $V \unlhd G_{\alpha}$ we have equality. Also if $g \in C_G(V)$, then $g$ also permutes the fixed points, hence similar $G_G(V) \le G_{\alpha}$. As $C_G(V)^h = C_G(V^h)$ we have $G_{\alpha} \in N_G(C_G(V))$ and $C_G(V) \cap G_{\alpha} \unlhd G_{\alpha}$, but as $V$ is the only normal subgroup we either have $C_G(V) \cap G_{\alpha} = G_{\alpha}$ or $C_G(V) \cap G_{\alpha} = V$, but the first case could be excluded as $V$ is obviously not central in $A_4$. So $C_G(V) \cap G_{\alpha} = V$ is the best I can get.