Note that \[ z_{n+1}=\frac{iz_n}{\overline{z}_n}=\frac{iz_n^2}{z_n\overline{z}_n}=\frac{iz_n^2}{|z_n|^2}.\]Since $ |z_0|=1 $, the sequence satisfies \[ z_1 = i z_0^2, \ z_2 = iz_1^2 = i\left(iz_0^2\right)^2 = -iz_0^4, \]and, in general, when $ k\ge 2 $, \[ z_k = -iz_0^{2^k}.\]Hence $ z_0 $ satisfies the equation $ 1 =-iz_0^{(2^{2005})} $, so $ z_0^{(2^{2005})} = i $. Because every nonzero complex number has $ n $ distinct $ n $th roots, this equation has $ 2^{2005} $ solutions. So there are $ \boxed{2^{2005}} $ possible values for $ z_0 $.