A matrix $A$ is normal if $AA^*=A^*A$. Suppose $(\lambda,x)$ is an eigenpair of $A$, i.e., $Ax = \lambda x$. Proof for a normal matrix $A$, $(\lambda,x)$ is an eigenpair of $A$ if and only if $(\bar{\lambda},x)$ is an eigenpair of $A^*$.
My try:
We have $Ax = \lambda x$ and want to show $A^*x=\bar{\lambda}x$. Taking the conjugate transpose we have $x^*A^* = \bar{\lambda} x^*$. Now we can write $x^*A^* A= \bar{\lambda} x^*A$, using the fact that $A$ is normal, we have $x^*A A^*= \bar{\lambda} x^*A$. Now I do not know how to get $A^*x=\bar{\lambda}x$.
Proving one direction is enough. Think in norm. Equality in normed space means zero in norm: $\Vert (A-\lambda I) x \Vert = 0$, then take transpose inside: $\Vert x^* (A^*- \lambda I) \Vert = 0$. Square both sides and apply the given condition that $A$ is normal to get $$ x^* AA^* x +\Vert \lambda\Vert^2 x^* x -\bar{\lambda} x^* A x - \lambda x^* A^* x.$$ Factorize this into $\Vert (A^*-I) x \Vert^2 = 0$.