Computing an expression involving characters Suppose that $G$ is a finite group, and let $\rho: G\to \text{GL}(V)$ be an irreducible representation with $\dim V = n$ (let's work over complex numbers, so $V$ is a $\mathbb{C}$-vector space). I will denote $\chi$ to be the character of $\rho$. (Recall that $\chi: G\to \mathbb{C}$ is defined by $\chi(s) = \operatorname{Tr}(\rho(s))$ for each $s\in G$).
Now, let's fix $t\in G$. I want to compute the following expression:
$$
(\ast) \ \ \ \ \ \ \ \ \ \ \sum_{s\in G} \chi(s^{-1}) \chi(t^{-1} s)
$$
I came across this sort of expression while working on Exercise 6.4 in Serre's Linear Representations of Finite Groups. Working backwards, I believe the answer should be $\dfrac{\# G}{n} \chi(t^{-1})$, where again $n$ is the dimension of the representation.  
So the question is: How do we compute the expression $(\ast)$? I have a hunch that we may need to use orthogonality of characters, but I have not been successful so far. 
 A: Denote by $\psi \in \mathbb{C}[G]$ the projector onto the irreducible representation $V$ with character $\chi$, which is given by the following formula:
$$ \psi = \frac{\dim V}{|G|} \sum_{s \in G} \chi(s^{-1}) s$$
It is not too hard to see that $\psi$ acts on the irrep $V$ by the scalar $1$, and on any non-isomorphic irrep by $0$. You can prove this by first showing that $\psi$ belongs to the centre of the group algebra, so by Schur's lemma $\psi$ must act on any irreducible representation by a scalar. You can find that scalar by taking the trace of $\psi$ for different irreps, and applying character orthogonality.
The expression you want to compute is, up to a scalar multiple, the trace of the operator $t^{-1} \psi |_V$, where I use the notation $x|_V$ for the image of $x \in \mathbb{C}[G]$ in $\mathrm{End}(V)$. $\psi|_V$ acts by the scalar $1$, and so $\psi|_V = \mathrm{id}_V$, and the operator $t^{-1} \psi |_V = t^{-1}|_V$. Then $\mathrm{tr}(t^{-1} \psi |_V) = \mathrm{tr}(t^{-1}|_V) = \chi(t^{-1})$ and from this your result follows.
