Calculation of trace of linear map $u\mapsto\frac{1}{|G|}\sum\limits_{g\in G} \chi(g^{-1})gu$

Let $$G$$ be a finite group, $$V$$ a $$\mathbb{C}G$$-module with character $$\chi$$. Let $$z:=\frac{1}{|G|}\sum\limits_{g\in G}\chi(g^{-1})g\in\mathbb{C}G$$. Let $$U$$ be an irreducible $$\mathbb{C}G$$-module with characer $$\psi$$. We define $$\zeta:U\to U$$ by $$\zeta(u)=zu$$. It is easy to show that $$\zeta$$ is $$\mathbb{C}G$$-homomorphism. By Schur's lemma, there exists $$\lambda\in\mathbb{C}$$ such that $$\zeta(u)=\lambda u$$. Then Tr$$_U(\zeta)=\lambda\dim(U)$$. Now my question asks to calculate Tr$$_U(\zeta)$$ in terms of $$\psi$$ and $$\chi$$. How can I do this?

By definition the trace of $$g$$ acting on $$U$$ is $$\psi(g)$$. So the trace of $$\frac1{|G|}\sum_{g\in G}\chi(g^{-1})g$$ acting on $$U$$ is $$\frac1{|G|}\sum_{g\in G}\chi(g^{-1})\psi(g).$$ This is an integer, the number of copies of the irreducible module $$U$$ within the module $$V$$.