This is an exercise from M.Isaac's book on character theory of finite groups (exercise 4.6, Pg 60):
Let $n \in {\mathbb{N}}$ and assume that $\chi^{(n)} \in {\mathrm{Irr}}(G)$ for every $\chi \in {\mathrm{Irr}}(G)$. Show that $G = H \times A$, where $A$ is abelian and ${\mathrm{g.c.d.}}(|H|, n) = 1$. Here $\chi^{(n)}(g) := \chi(g^n)$.
The first hint was: Let $d = {\mathrm{g.c.d.}}(|G|, n)$. Show that it is no loss to assume that ${\mathrm{g.c.d.}}(|G|/d, n) = 1$. [The conclusion is sort of obvious. Surely a priori it is difficult to find a subgroup $H$ with ${\mathrm{g.c.d.}}(|H|, n) = 1$]
The second hint was: Let $A = \bigcap_{\chi \in {\mathrm{Irr}}(G)} {\mathrm{ker}}(\chi^{(n)})$. Show that $A = \{ g \in G ~:~ g^n = 1 \}$ and $|A| = d$.
The first statement follows from the definition of $\chi^{(n)}$ itself. ${\textbf{ It is not so clear why $A$ must be abelian}}$. It is easy to realise $A$ being kernel of $\rho^{(n)}$, where $\rho$ is the regular character. The problem is many of the $\chi^{(n)}$ will coincide for different $\chi \in {\mathrm{Irr}}(G)$.
The third hint was: Let $H = \bigcap \left\{ {\mathrm{Ker}}(\chi) ~:~ \chi \in {\mathrm{Irr}}(G), \chi^{(n)} = 1_{G} \right\}$. Show that $[G : H] = d$.
Assuming the previous step, one can apply Schur-Zassenhaus theorem to conclude this. ${\textbf{Is there any direct way?}}$