Solution:

Answer: any $C$ such that $\angle ACB=180^{\circ} r / s$, with $r$ and $s$ relatively prime integers and $s$ not a power of $2$.

Let $\angle AC_{n}B = x_{n}$, where the angle is measured clockwise, so that $x_{n}$ is positive on one side of $AB$ and negative on the other side. Then $x_{n}$ uniquely identifies $C_{n}$ on the perpendicular bisector.

We have $x_{n+1} = 2 x_{n}$. To make this work in all cases we have to take it mod $180^{\circ}$ (so that if $AC_{n}B$ is obtuse, then $C_{n+1}$ lies on the other side of $AB$). If $x_{n}$ is eventually periodic then $x_{m+1} = x_{n+1}$, for some $n > m$, so $(2^{n} - 2^{m}) x_{1} = 0 \bmod 180$. Hence $x_{1} = 180 r / s$ for some relatively prime integers $r, s$. Also $s$ cannot be a power of $2$ for then we would have $x_{k} = 180 r$ for some $k$, in which case the sequence would terminate rather than be periodic.

Conversely, suppose $x_{1} = 180 r / s$, with $r$ and $s$ relatively prime and $s$ not a power of $2$. Then $x_{n+1} = 180 \cdot 2^{n} r / s$ cannot be $0 \bmod 180$, so the sequence does not terminate. Put $s = 2^{b} c$, with $c$ odd. Let $d = \varphi(c)$, where $\varphi(m)$ is Euler's phi function, so that $2^{d} = 1 \mod c$. Then $x_{b+1} = 180 r / c \bmod 180$ and $2^{b+d} = 2^{b} \bmod c$, so $x_{b+d+1} = 180 r / c \mod 180$. Hence the sequence is periodic.