Let $S$ be a non-trivial simple group and suppose $S \trianglelefteq G$ if $C\_G(S)=1$ then $S$ is characteristic in $G$. To prove this let $\phi$ be an automorphism of $G$ and note that the intersection $S\cap \phi(S)$ can't be trivial since otherwise $S$ commutes with $\phi(S)$ in $G$. Therefore since both $S$ and $\phi(S)$ are simple $S =S \cap \phi(S) = \phi(S)$.


Is there a non-trivial perfect centerless group $P \trianglelefteq G$ such that $C\_G(P)=1$ which is not a characteristic subgroup of $G$?



Perfect centerless normal subgroups
Let $S$ be a finite simple group and $V$ a faithful absolutely irreducible module for $S$. Then $W = V \otimes V$ is a faithful irreducible module for $S \times S$.


Let $G = W \rtimes (S \times S)$ be the corresponding semidirect product of $W$ with $S \times S$. Then $G$ has two normal subgroups of the form $W \rtimes S$, and they are both perfect with trivial centralizer in $G$, but they are not characteristic in $G$, because there is an automorphism that interchanges them.


For example, we could take $S = A\_5 \cong {\rm SL}(2,4)$ and $V$ the natural module of order $4^2$, giving a group $G$ of order $2^8 \times 60^2=921600$. (I checked that one on the computer.)


