Let $G$ be a permutation group on $\Omega$ with $G = VN$, where $V \cong C_2 \times C_2$ (the four-group) and $N$ has odd order with some prime divisor $>3$. Suppose $N$ is a regular normal subgroup.
There exists some prime $p \ne 3$ and a Sylow $p$-subgroup $S$ of $N$ which is left invariant by $V$.
Do you know any proof of this fact?
$\newcommand{\Size}[1]{\lvert #1 \rvert}$I think the following more general result is true.
Let $G = P N$ be a group, where $P$ is a $p$-group for some prime $p$, $N$ is a $p'$-group, and $N$ is normal in $G$. Then for each $q$ dividing the order of $N$, there is a $q$-Sylow subgroup of $N$ normalized by $P$.
Let $\Delta$ be the set of $q$-Sylow subgroups of $N$, and let $Q$ be one of these subgroups. The size of $\Delta$ is $\lvert N : N_{N}(Q) \rvert$, a $p'$-number.
Now $P$ acts by conjugation on $\Delta$. But not all orbits can have an order which is a multiple of $p$, otherwise the order of $\Delta$ would be divisible by $p$.
So there is an orbit $\{ Q \}$ of length 1, and thus $P$ leaves $Q$ invariant.
Or alternatively, fix a $q$-Sylow subgroup $Q$ of $G$. Then the number of $q$-Sylow subgroups is $$ \Size{N : N_{N}(Q)} = \Size{N : N \cap N_{G}(Q)} = \frac{\Size{N \cdot N_{G}(Q)}}{\Size{N_{G}(Q)}} = \Size{G : N_{G}(Q)}, $$ so that $G = N \cdot N_{G}(Q)$. It follows that $N_{G}(Q)$ contains a $p$-Sylow subgroup of $G$, that is, a conjugate of $P$. Back-conjugating you get that $P$ normalizes a $q$-Sylow subgroup of $N$.