If you have a wave function ##\psi(t,\vec{x})## as the solution of the time-dependent Schrödinger equation, you can easily show that the conjugate function solves the conjugate Schrödinger equation, but the conjugate Schrödinger equation looks pretty much like the Schrödinger equation, except that the sign of the term with the time derivative changes, because of the ##\mathrm{i}## in front of it. The Schrödinger equation reads
$$\mathrm{i} \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x})=\left [-\frac{\Delta}{2m}+V(\vec{x}) \right ] \psi(t,\vec{x}).$$
Now take the conjugate complex of this equation. The only change is the sign on the left side of the equation and that the conjugate wave function occurs:
$$-\mathrm{i} \partial_t \psi^*(t,\vec{x})=\left [-\frac{\Delta}{2m}+V(\vec{x}) \right ] \psi^*(t,\vec{x}).$$
This shows that complex conjugation has to do with the "time-reversal transformation", because the changed sign in front of the time derivative can be interpreted as following the time evolution backwards. Indeed, it's immediately clear from the equation, that the solution of the complex conjugate equation is not a solution of the original Schrödinger equation, but obviously ##\psi^*(-t,\vec{x})## is.
Indeed, it turns out that the time-reversal transformation is realized as an anti-unitary operator rather than a unitary one. For details, see a good QT textbook, e.g., Sakurai&Napolitano.
