In Section 3.2 of this textbook (pg. 91), Ramakrishnan considers the function $\phi(s) = (\rho(s)[x],x)$, where $(\cdot,\cdot)$ denotes an inner product and $\rho$ denotes a representation of a top. group $G$ in the space of unitary operators of a Hilbert space $H$. So, $s \in G, x \in H$.
He writes...
$$\sum_{i,j=1}^n \phi(s_j^{-1}s_i)z_j\bar{z_i} = \sum_{i,j=1}^n (\rho(s_j^{-1}s_i)[x],x)z_j\bar{z_i} =_{(2)} \sum_{i,j=1}^n(\rho(s_i)[x],\rho(s_j)[x])z_j\bar{z_i}$$
$$=_{(3)} (\sum_{i=1}^n \rho(s_j)(z_jx),\sum_{j=1}^n \rho(s_i)(z_ix))$$
Question (2) is because $\rho(s_j)$ is unitary + can be cancelled with its inverse since its a representation. However, I am unable to see why equality (3) is true.
My reasoning so far... I know that, for a complex inner product, $(ax,by)=(x,y)a\bar{b}$ for complex constants $a,b$. Also, that it is conjugate symmetric, i.e. $(x,y)=\overline{(y,x)}$ (Ramakrishnan uses this assumption here, at least). Therefore, I can see the following:
$$(\rho(s_i)[x],\rho(s_j)[x])z_j\bar{z_i} = \overline{(\rho(s_j)[x],\rho(s_i)[x}]) z_j\bar{z_i} = \overline{(\rho(s_j)[x],\rho(s_i)[x]) \bar{z_j}z_i}$$
But this is not the same as what we desire.
No answer was found, so I am not sure