If there is a function $H(e^{j\omega})$. Then will its conjugate be $H(e^{-j\omega})$?Why?
What you say only holds true for real signals. Let's prove it by the definition of Fourier transform:
$$
H(e^{j\omega}) = \mathcal{F}[h(t)] = \int_{-\infty}^{\infty} h(t) e^{-j\omega t} dt
$$
and $H^*(e^{j\omega})$ should be
$$
H^*(e^{j\omega}) = \int_{-\infty}^{\infty} h^*(t) e^{j\omega t} dt
$$
thus
$$
H^*(e^{-j\omega}) = \int_{-\infty}^{\infty} h^*(t) e^{-j\omega t} dt = \mathcal{F}[h^*(t)] 
$$
If $h(n)$ is real, $\mathcal{F}[h^*(t)] = \mathcal{F}[h(t)] $, we can derive that
$$
H^*(e^{-j\omega}) = H(e^{j\omega}) 
$$
Take conjungate on both sides of the equation above, we have
$$
H(e^{-j\omega}) = H^*(e^{j\omega}) 
$$