Does $Ax=x$ imply $A^* x=x$, if $A^*$ is the conjugate transpose of $A$?

No, take the matrix

$$\begin{pmatrix} 1 & 1\\ 0 & -1\end{pmatrix}$$

which has $x=(1,0)^T$ as an eigenvector with eigenvalue 1. Yet $A^*x=(1,1)^T\neq x$.

The easiest example would be to consider the rank one matrix $$A = xy^\top$$ Then $$Ax = (xy^\top) x= x(y^\top x) = x\lambda = \lambda x$$ and $$ A^* x = (\bar{y} \bar{x}^\top) x =\bar{y} (\bar{x}^\top x )= \bar{y} k = k\bar{y}$$