Note that:
$$z = re^{i\phi}, z^5 = r^5 e^{i5\phi}~\text{and}~\bar{z} = re^{-i\phi},$$
where $r \geq 0$. Therefore:
$$z^5 = \bar{z} \Rightarrow
\begin{cases}
r^5 = r\\
5\phi + 2k\pi= -\phi + 2h\pi
\end{cases},$$
where $k, h \in \mathbb{Z}.$
The previous system can be rewritten as:
$$
\begin{cases}
r(r^4-1) = 0\\
6\phi = 2s\pi
\end{cases},$$
where $s = k-h \in \mathbb{Z}.$
The first equation has $3$ distinct roots: $r=-1$, $r=0$ and $r=1.$ Of course, $r=-1$ should be discarded. This means that for $r=0$, $z = 0$ is a solution, which obviously does not depend on the phase $\phi.$ Moreover, for $r=1$, the phase is important. Solving the second equation, we get:
$$\phi = \frac{s\pi}{3},$$
and hence $z = e^{\frac{is\pi}{3}}$ for $s \in \mathbb{Z}$, together with $z=0$ represent the solution of the equation.
Best Answer
If $z^2=\bar z$ then taking magnitude gives $|z|^2=|z|$ so $|z|=0,1$. The case $|z|=0$ gives $z=0$.
If $|z|=1$, then the equation $z^2=\bar z=z^{-1}$ so $z^3=1$. This gives the remaining 3 solutions which are the third roots of unity.