Let $\chi$ denotes (complex) irreducible character of a finite group $G$. Define $\chi^{(n)}$ to be the function from $G$ to $\mathbb{C}$ by $\chi^{(n)}(g)=\chi(g^n)$.
Problem: For fixed $n>0$, if $\chi^{(n)}$ is irreducible whenever $\chi$ is irreducible, then $G=H\times A$ with $A$ abelian and $(|H|,n)=1$.
This question has one hint the following:
(Hint:) Let $d=\gcd\left(\left|G\right|,n\right)$. Show that it is no loss to assume that $gcd(|G|/d,n)=1$.
I couldn't get any direction to prove the hint. Can one elaborate the hint a little?
Note. (This is not the duplicate of this question.)
Notice that if we take a sufficiently large $k \in \mathbb{Z}$ and set $d' = \gcd(|G|,n^k)$, we have that $\gcd(|G|/d',n^k) = 1$. We may replace $n$ with $n^k$ since if $\chi^{(n)} \in \mbox{Irr}(G)$ for each $\chi \in \mbox{Irr}(G)$, then by induction, $\chi^{(n^k)} \in \mbox{Irr}(G)$ for each $\chi \in \mbox{Irr}(G)$.