Let $G$ be a finite transitive permutation group on $\Omega$. Suppose that $G_{\alpha}$ is a $3$-group and that $|\mbox{fix}_{\Omega}(G_{\alpha})| = 3$ and that $G$ contains a regular normal subgroup, i.e. we have $G = N \rtimes G_{\alpha}$ and $N$ is transitive. Also assume that $|\mbox{fix}_{\Omega}(g)|\le 3$ for each nontrivial $g \in G$.
Set $H := G_{\alpha}$ and take $P \in \mbox{Syl}_p(G)$ such that $H \le P$. As $H$ acts on $N$ by conjugation similar to its action on $\Omega$, we have $|C_N(H)| = 3$, hence $3 \in \pi(N)$ and $H \ne P$. By properties of $p$-groups we have $H < N_G(H) \cap P \le P$ and as $N_G(H)$ acts on the three fixed points $|N_G(H) : H| \le 3$. This gives $N_G(H) = N_P(H)$ and assume $|P : H| = 3$ (this is the case my question is concerned with, the case $|P : H| > 3$ should not bother us here) from now on. So with $H \le P$ $$ |H| = |G/N| = |PN/N| = |P/P\cap N| $$ and we find that $|P\cap N| = 3$ (and this is a Sylow subgroup of $N$). As $H \le P$ we have that $H$ normalizes $N_N(P\cap N)$ and $C_N(P\cap N)$ and hence acts as an automorphism on their quotient, which is isomorphic to a subgroup of $\mbox{Aut}(P\cap N)$, which has order $2$.
Now we want to prove that $N_N(P\cap N) = C_N(P\cap N)$. So assume to the contrary that $N_N(P\cap N) \ne C_N(P\cap N)$, then $|N_N(P\cap N) / C_N(P\cap N)| = 2$ and hence $H$ centralizes this quotient.
From the above it follows that $C_N(H)$ has even order. Hence a contradiction to $|C_N(H)| = 3$, so that $N_N(P\cap N) \ne C_N(P\cap N)$ is not possible.
Why does it follow that $C_N(H)$ has even order if $H$ centralizes $N_N(P\cap N) / C_N(P\cap N)$? This just means that $[n,h] \in C_N(P\cap N)$ for each $n \in N_N(P\cap N)$ and $h \in H$, or that $n^h = nc$ for some $c \in C_N(P\cap N)$, but I do not see how is this connected to the order of $C_N(H)$. If we would have $N_N(P\cap N) \le C_N(H)$ then it would follow, but I am not sure if this holds?
Just some remarks. From $|\mbox{fix}_{\Omega}(G_{\alpha})| = 3$ by transitivity this holds for all stabilizers and they are all three-point stabilizers. This also gives that if $G_{\alpha} \ne G_{\alpha}^g$, then $G_{\alpha}\cap G_{\alpha}^g = 1$ as every nontrivial element has at most three fixed points. And so every nontrivial element has either no fixed point or exactly three fixed points. Also observe that this already implies $N_G(H) \ne H$ and more $|N_G(H) : H| = 3$, for if $G_{\alpha} = G_{\beta} = G_{\gamma}$ and $\alpha^g = \beta$, then $G_{\alpha}^g = G_{\beta} = G_{\alpha}$, hence $g \in N_G(H)$ and $g$ permutes the fixed points $\{\alpha, \beta, \gamma\}$ and could not fix $\gamma$ or $\beta$ (as then it would fix $\alpha$ too by equality of the stabilizers), hence $\beta^g = \gamma$ and as $g^2g^{-1} = g \notin H$ the three cosets $H, gH, g^2H$ are distinct.