I'm reading through a proof where $\Psi:G\rightarrow GL(V)$ and $\Phi:G\rightarrow GL(U)$ are representations of a finite group $G$, $a\in U$, $b\in V$, and $R:U\rightarrow V$ is a linear function. Then there is a line in the proof that goes: $$\sum_{g\in G}\langle b,\Psi(g^{-1})R\Phi(g)a\rangle=\sum_{g\in G}\langle\Psi(g)b,R\Phi(g)a\rangle$$ I thought in general that for an operator $A$, $\langle b,Aa\rangle=\langle A^*b,a\rangle$, so it looks like that line is assuming $\Psi(g^{-1})=\Psi(g)^*$, i.e., $\Psi(g)$ is unitary.
I can see that since $g^n=e$ for some $n$, then $\Psi(g)^n=I$, so $det(\Psi(g))$ is an $n$th root of unity, but as far as I can tell that's not a sufficient condition to conclude that $\Psi(g)$ is unitary.
Does $\Psi(g)$ need to be unitary when $G$ is finite?
No, it does not. For example, the representation of the $2$-element group $\{e,a\}$ by $$ 1 \to \pmatrix{1 & 0\cr 0 & 1\cr},\ a \to \pmatrix{1 & 1\cr 0 & -1\cr} $$