Let $G$ be a finite transitive permutation group, suppose $p$ does not divide $G_{\alpha}$ and that $1 \ne P \unlhd G$ is a normal $p$-subgroup of $G$. Let $\Delta := \alpha^P$ be an orbit of $P$ and suppose that $g \in G_{\alpha}$ just has the unique fixed point $\alpha$ on $\Delta$ and acts semi-regular on $\Delta \setminus \{\alpha\}$, i.e. every orbit of size $> 1$ has size $|g|$. Then $|P| = |\Delta| = 1 + k\cdot |g|$ for some $k \in \mathbb N$.
Why does this implies that $g$ fixes a point on every other orbit of $P$ that it stabilizes?
If $\Gamma := \beta^P$ would be another orbit of $P$ such that $\Gamma^g = \Gamma$, then $g \in G_{\beta}P$. But why has $g$ to fix some point in $\Gamma$? Just the numerical relations do not seem to imply that, for example if $|P| = 7$ and $|g| = 6$, then the orbits of $g$ on $\Gamma$ might have sizes $2,2,3$ and so no fixed point, but on $\Delta$ we might have $7 = 1 + 6$ for the orbit sizes.
With Derek's hints I collect what is written in the comments. The notation as in the question.
Lemma: If $N \unlhd G$ is a regular normal subgroup, then $G_{\alpha}$ acts similar on $\Omega$ as on $N$ by conjugation.
Proof: Fix $\alpha \in \Omega$ and define $\lambda : N \to \Omega$ by $\lambda(x) := \alpha^x$. Then for $g \in G_{\alpha}$ we have $\lambda(x^g) = \alpha^{g^{-1}xg} = \alpha^{xg} = \lambda(x)^g$. $\square$
Corollary: If $N \unlhd G$ and $g$ fixes exactly $k > 0$ points, we have $|C_P(g)| = k$.
Proof: Choose some $g \in G_{\alpha}$, then $x \in C_P(g)$ iff $\lambda(x)^g = \lambda(x)$, hence the claim follows. $\square$
Now for the situation in the question, the assumptions give $C_P(g^i) = 1$ for nontrivial $g^i$. As if $g^i \ne 1$ the subgroup $C_P(g^i)$ acts on the fixed points of $g^i$ on $\Delta$. But by assumption $\alpha$ is the only fixed point of $g^i$ on $\Delta$ and hence every $y \in C_P(g^i)$ has to fix $\alpha$ too, but as $P_{\alpha} = 1$ this implies $C_P(g^i) = 1$. But this could also be seen with the Lemma by the assumption about the orbit structure on $\Delta$.
Now let $\Gamma$ be a different orbit of $P$ such that $\Gamma^g = \Gamma$. If $g$ fixes no point of $\Gamma$, then as $|\Gamma| = 1 + k|g|$ it could not act semi-regular on $\Gamma$, hence we have some $g^i \ne 1$ that fixes a point of $\Gamma$. Now look at the cycle structure induced by $g$ on $\Gamma$. The different orbits correspond to distinct cycles and by the above we have some cycle of length $i$ with $1 < i < |g|$; and we see that $g^i$ must fix at least $i$ points, i.e. all the ones in the cycle of length $i$. Also $P$ acts regular on its $P$-orbit $\Gamma$; and together with the Lemma the action of $g^i$ on $P$ by conjugation is equivalent to its action on $\Gamma$, hence $|C_P(g^i)| = i > 1$. A contradiction. So $g$ itself must fix a point in $\Gamma$.