Define Hol(G) as $\hat{G}\rtimes Aut(G)$, where $\hat{G}$ means the group of right multiplication induced by $G$.
The problem is to prove Hol($G$) is primitive when $G$ is characteristically simple.
Since $\hat{G}\le$Hol($G$), we know Hol($G$) acts transitively on $G$.
So I want to use a well-known result, which states that a transitive group is primitive if and only if its stabilizer (of any point) is a maximal subgroup.
Since Hol($G$) is transitive, so it suffices to consider the stabilizer of identity of $G$, denoting the identity as $\alpha$.
Since $\hat{G}$ acts regular on $G$, we know $\hat{G}_{\alpha}$ is $\{1\}$. And so Hol($G$)$_\alpha$ is $Aut(G)$.
But I don't know how to use the condition that $G$ is characteristically simple to prove $Aut(G)$ is maximal in Hol($G$).
Any help will be appreciated. Thanks!
This is straightforward. If ${\rm Aut}(G)$ is not maximal in ${\rm Hol}(G)$, then it is strictly contained in a proper subgroup of $G$, which must have the form $N \rtimes {\rm Aut}(G)$ for some $N$ with $1 < N < G$. But then $N$ is a subgroup of $G$ that is invariant under the action of ${\rm Aut}(G)$, contradicting the assumption that $G$ is characteristically simple.
Note that the converse is true: if $G$ if $G$ is not characteristically simple, then $G \rtimes {\rm Aut}(G)$ is imprimtiive.