Cheng & Li gives the following problem:

Let $\psi_1$ and $\psi_2$ be the bases for the spin-1/2 representation of $su(2)$ and that for the diagonal operator $T_3$, \begin{align} T_3\psi_1 &= \frac{1}{2} \psi_1 \\ T_3\psi_2 &= -\frac{1}{2}\psi_2 \end{align} What are the eigenvalues of $T_3$ acting on $\psi_1^*$ and $\psi_2^*$ in the conjugate representation?

I originally thought this problem was trivial, just take the complex conjugate of both sides and use the fact that $T_3$ is real valued to get that $T_3\psi_1^* = \frac{1}{2}\psi_1^*$, but this is wrong.

If we start from the arbitrary transformation $\psi'_i = U_{ij}\psi_j$ and complex conjugate both sides, we get ${\psi'}_i^* = U_{ij}^* \psi_j^*$. But for traceless Hermitian matrices such as $U$, there exists an $S \in su(2)$ such that $S^{-1}US = U^*$, and so, writing the previous equation in matrix form: \begin{align} \psi'^* = (S^{-1}US)\psi^* \implies S\psi'^* = U(S\psi^*) \end{align} So $S\psi^*$ transforms as $\psi$. It turns out that in the Pauli representation that $S = i\sigma^2$, and so: \begin{align} T_3\left(\begin{matrix} \psi_2^* \\ -\psi_1^*\end{matrix}\right) = \left(\begin{matrix} 1/2 & 0 \\ 0 & -1/2\end{matrix}\right) \left(\begin{matrix} \psi_2^* \\ -\psi_1^* \end{matrix}\right) \end{align}

What I don't understand is why we couldn't just take the complex conjugate of both sides? Is this quantity $\psi^*$ not the traditional "algebraic" complex conjugate of $\psi$? If so, why could we complex conjugate $\psi'_i = U_{ij}\psi_j$ to get $\psi^*$? I feel like I thought I understood the conjugate representation but I clearly don't and I would appreciate any help understanding it.