I am preparing to an exam in representations of finite groups. I am trying to tackle a problem regarding a characterization of irreducible characters: Let $f$ be a complex-valued function on a finite group $G$. Then: $$\forall x,y\in G: f(x)f(y) = \frac{f(1)}{|G|}\sum_{z\in G} f(yzxz^{-1}) \leftrightarrow f \text{ is proportional to an irreducible character}$$
I have made some progress in $\rightarrow$: if $f$ is identically 0, then RHS is implied. Otherwise, it is easy to see that $f$ is a class function (choose $y$ such that $f$ doesn't vanish on it, and notice that $\{yzxz^{-1} | z \in G \}$ is invariant under conjugation of $x$. I thought about proving $\leftarrow$ and then writing $f$ as a combination of irreducible characters.
I also noticed the following: $f(1)$ is the degree of the representation when $f$ is a character (and it appears often in orthogonality relations). Also, the projection on an isotypical component (corresponding to an irreducible representation $\pi$) of a representation $\rho$ is $\frac{\chi(1)}{|G|}\sum_{g} \overline{\chi(g)}\rho(g)$ (where $\chi$ is $\pi$'s irreducible character).
To sum it all: I have some directions, but haven't used orthogonality relations explicitly, for example. Can you help?
EDIT: Some more advances: I've noticed that if $\phi$ is a class function, the equation is equivalent to $$\sum_{z\in G} \phi(x) \phi(z^{-1} y z) \sim \sum_{z \in G} \phi(x (z^{-1}yz))$$ which looks nicer ($\sim$ implies that LHS is proportional to RHS). By putting $y=1$ it is easy to deduce that $\sum_{z\in G} \phi(x) \phi(z^{-1} y z) = \phi(1) \sum_{z \in G} \phi(x (z^{-1}yz))$ (i.e., $\phi(1)$ is the ratio). Some directions in $\leftarrow$: if $\rho$ is an irreducible representation with a simple character $\chi$, then we can let $\rho ' = \sum_{z \in G} \rho_{z^{-1}yz}$. The equation becomes $Tr(\rho_x)Tr(\rho ') = \chi(1) Tr(\rho_x \rho ')$. The product of characters might suggest use of tensor product.