Is that true that the normalizer of $H=A \times A$ in symmetric group cannot be doubly transitive? Is that true that the normalizer of $H=A \times A$ in the symmetric group cannot be doubly transitive where $H$ is non-regular, $A \subseteq H$ and $A$ is a regular permutation group?

Let the group $A$ act regularly on a set $X$. Then the centralizer of $A$ in the symmetric group $S_X$ is a regular group $A'$ isomorphic to $A$. (The centralizer of the (image of the) left regular representation of $A$ is the right regular representation.) Note that $A \cap A' = Z(A)$, so we only get $A \times A$ acting in the manner described if $Z(A)=1$. So let's assume that $Z(A)=1$.
Let $G=N_{S_X}(H)$ with $H=A \times A$ and suppose that $G$ is $2$-transitive. Now the two direct factors of $H$ are fixed or interchanged by elements of $G$, so  the normalizer $N$ of the two factors has index $2$ in $G$. Then, for any $x \in X$,
$N_x \unlhd G_x$ with $|G_x:N_x| \le 2$, so $N_x$ either acts transitively, or it has two orbits of equal size on $X \setminus \{x\}$.
Now since $A$ acts regularly on $X$, for each $y \in X$ there is a unique $a \in A$ with $a(x)=y$, and so we can identify $A$ with $X$, with $x \in X$ corresponding to the identity element $1 \in A$. If we do that, then the action of $N_x$ on $X$ corresponds to the conjugation action of $N_x$ on $A$.
So, if $N_x$ is transitive on $N \setminus \{x\}$, then all elements of $A \setminus \{1\}$ are conjugate under $N_x$, so they all have the same order, which must be a prime, so $A$ is a $p$-group, contradicting $Z(A) = 1$.
Otherwise $N_x$ has two orbits of the same size on $X \setminus \{x\}$, and so (since $A$ cannot be a $p$-group), $A$ must have equal numbers of elements of orders $p$ and $q$ for distinct primes $p$ and $q$. So $|A|=p^aq^b$ for some $a,b > 0$. Also, since the elements of orders $p$ and $q$ in $A$ are all conjugate under $N$, the Sylow $p$- and $q$-subgroups of $A$ must be elementary abelian, so all conjugacy classes of $p$-elements have order a nonzero power of $q$ while those of order $q$ have order a nonzero power of $p$. Hence $(p^aq^b-1)/2$ is divisible by $p$ and $q$, which is clearly impossible.
So $G$ cannot be $2$-transitive.
