Let $z_1 = r_1 e^{i\theta_1}, \qquad z_2 = r_2 e^{i\theta_2}$ where $\theta_1 = \arg z_1$ and $\theta_2 = \arg z_2$.

Given $\overline{z_1} = i\,\overline{z_2}.$ Taking arguments on both sides: $\arg(\overline{z_1}) = \arg(i\overline{z_2}).$ Now, $\arg(\overline{z_1}) = -\theta_1, \qquad \arg(\overline{z_2}) = -\theta_2, \qquad \arg(i)=\frac{\pi}{2}.$ Hence, $-\theta_1 = \frac{\pi}{2} - \theta_2 \pmod{2\pi}.$ So, \theta_1 = \theta_2 - \frac{\pi}{2} \pmod{2\pi}. \tag{1}

Also given $\arg\left(\frac{z_1}{\overline{z_2}}\right)=\pi.$ Therefore, $\arg(z_1)-\arg(\overline{z_2}) = \pi \pmod{2\pi}.$ Since $\arg(\overline{z_2})=-\theta_2$, \theta_1 + \theta_2 = \pi \pmod{2\pi}. \tag{2}

Substitute (1) into (2): $\left(\theta_2 - \frac{\pi}{2}\right)+\theta_2 = \pi \pmod{2\pi}.$ Thus, $2\theta_2 - \frac{\pi}{2} = \pi \pmod{2\pi}$ $2\theta_2 = \frac{3\pi}{2} \pmod{2\pi}$ $\theta_2 = \frac{3\pi}{4} \pmod{\pi}.$ So possible principal values are $\theta_2 = \frac{3\pi}{4} \quad \text{or} \quad -\frac{\pi}{4}.$ Neither matches options A or B.

Now find $\theta_1$ using (1): $\theta_1 = \theta_2 - \frac{\pi}{2}.$ If $\theta_2 = \frac{3\pi}{4}$, then $\theta_1 = \frac{\pi}{4}.$ If $\theta_2 = -\frac{\pi}{4}$, then $\theta_1 = -\frac{3\pi}{4}.$

Check which option is consistent with the principal argument condition $\arg\left(\frac{z_1}{\overline{z_2}}\right)=\pi.$ Since the argument is specified as $\pi$ (not $-\pi$), take $\theta_1+\theta_2 = \pi.$ This is satisfied by $\theta_2 = \frac{3\pi}{4}, \qquad \theta_1 = \frac{\pi}{4}.$

Therefore, $\arg z_1 = \frac{\pi}{4}.$ So the correct option is C.