= Solution
The orbits of $N_\alpha\trianglelefteq G_\alpha$ on $\Omega\setminus\{\alpha\}$ have a common size $r$, because $G_\alpha$ is transitive there. Suppose $N$ has a <block system> of block size $b$, $1<b<n$. Since $r\mid n-1$ and $b\mid n$, $\gcd(r,b)=1$. Let $B$ be the block of $\alpha$ and $C\ne B$ another block. The union $N_\alpha C$ is a union of blocks and of r-element point-orbits, so its size is divisible by $br$; it is at most $br$. Equality means each of the b point-orbits meets $C$ once. Thus $N_{\alpha\beta}$ fixes $C$ pointwise for $\beta\in C$. Interchanging $\alpha,\beta$ also makes it fix $B$. All two-point stabilizers have order $|N_\alpha|/r$. Therefore $N_{\alpha\beta}=N_{\alpha\gamma}$ for every $\gamma\in B\setminus\{\alpha\}$. Varying $\beta$ outside $B$ shows this same <subgroup> fixes every point, so it is trivial. We have proved <uniform subdegrees force an imprimitive action to be Frobenius>: no nonidentity element of $N$ fixes two points.
Write $|N_\alpha|=h$. The distinct <point stabilizers> meet only at $1$, and $|N|=nh$. Thus the set $D$ of fixed-point-free elements of $N$ has size
$$
|D|=nh-\bigl(1+n(h-1)\bigr)=n-1.
$$
Count triples $(\alpha,\beta,x)$ with $x\in D$ and $x\alpha=\beta$: there are $n(n-1)$, exactly the number of ordered distinct pairs. <Conjugation> by the <two-transitive> <group> makes every pair occur, hence each pair has a unique such $x$. It follows that $G$ conjugates all elements of $D$ transitively.
Also $h\mid n-1$, since $N_\alpha$ acts freely on the remaining points. For each prime $p\mid n$, <Cauchy's theorem for finite groups> supplies an order-p element of $N$; it cannot fix a point, since $p\nmid h$. Conjugacy of $D$ forces all these primes to coincide, so $n=p^d$ and every element of $D$ has order $p$. A <Sylow subgroup> $P$ of $N$ has order $n$, because $p\nmid h$. Every nonidentity element of $P$ is fixed-point-free, so $P=\{1\}\cup D$. This uniquely described <subgroup> is normal in $G$. Minimal normality of $N$ gives $P=N$, and $|N|=n$. \b[Therefore $N$ is regular.] The proof establishes closure through a <Sylow subgroup>; it does not assume that a set of derangements is automatically a <subgroup>.
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