As anon points you out, try to use $\overline{z\cdot w} = \overline{z}\cdot \overline{w}$ and simplify notation.
For instance, you could just write vectors in $\mathbb{C}^2$ like this: $(z_1, z_2)$, with $z_1, z_2 \in \mathbb{C}$.
Second, I would try to convince myself that that product/conjugacy rule works too for products of matrices and vectors. That is:
$$
\overline{A\cdot v} = \overline{A}\cdot \overline{v} \ ,
$$
for, say, $A$ a $2\times 2$ complex matrix and $v\in \mathbb{C}^2$. What would happen if $A$ had real coefficitients?
Finally, I would write the equality I already know, namely
$$
A \cdot Y_0 = \lambda Y_0 \ .
$$
Staring at it should force inspiration to come. :-)
As $Y_0$ is an eigenvector you have $AY_O = \lambda Y_0$. Now take the conjugates of both sides, which for matrix means conjugating all entries you have:
$$\overline{AY_0} = \overline{\lambda Y_0} \iff \bar{A }\bar{Y_0} = \bar{\lambda} \bar{Y_0} \iff A\overline{Y_0} = \bar{\lambda}\bar{Y_0}$$
Therefore $\bar{\lambda}$ is an eigenvalue and $\overline{Y_0}$ is an eigenvector corresponding to it.
Best Answer
It is not true in general.
Take $A=\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix}$, then with $v^*=(1,1)$ we have $v^* A = v^* $, but $A v = (2,0)^T \neq v$.