what symmetry does system function H(z) have if h[n] is real?
it's because $e^{j\omega}$ is such a signal that has the property:

$$ (e^{j\omega})^* = e^{-j\omega} = \frac{1}{e^{j\omega}}$$

then going back from the DTFT to Z-transform is not obvious. So it's better to start from definition of Z-transform to define the symmetry:

$$ X(z) = \sum_{n} x[n] z^{-n} $$

and apply this to a signal which is real; $x[n] = x^*[n]$

$$ x[n] \xrightarrow{ Z-transform} X(z) \implies x^*[n] \xrightarrow{ Z-transform} X^*(z^*) $$

and conclude for real $x[n]$ :
$$ X(z)  = X^*(z^*) $$ 

Derivation:

$$\begin{align}
\mathcal{Z} \{x^*[n] \} &amp;= \sum_n x^*[n] z^{-n} \\ \\
&amp; = (\sum_n x[n] (z^*)^{-n})^* \\ \\
&amp; = X^*(z^*) \\
\end{align}
$$

As you can see, on the unit circle where $z = e^{j \omega}$, this becomes:

$$ X(e^{j\omega}) = X(e^{-j\omega})^* $$
the conjugate-symmetry property...