Solution:

We shall prove that the first player wins if and only if the exponent of $2$ in the prime decomposition of $n-3$ is odd.

Since the game is identical for both players, has finitely many possible states and always terminates, we can label the possible states Wins or Losses according as whether a player faced with that position has a winning strategy or not. A state is a Win if and only if there is some legal move taking the state to a Loss, and a state is a Loss if and only if all moves take that state to a Win (including the case where there are no legal moves).

Lemma. Any configuration in which the triangle formed by the three counters is not isosceles is necessarily a Win.

Proof. Label the positions of the counters $X, Y, Z$ so that the arc $YZ$ of the circumcircle is shortest and the $\operatorname{arc} ZX$ is longest. Begin by moving the counter at $Z$ around the polygon on the arc $YZX$ until it forms an isosceles triangle $XYZ'$ with apex at $Y$ (note that the $\operatorname{arc} XY$ is less than half the circle, so that $Z$ does not jump over the counter at $X$). If this configuration is a Loss, we are done.

If instead this configuration is a Win, then the counters can be moved legally from triangle $XYZ'$ to reach a losing state. This cannot involve the counter at $Y$, so by symmetry a Loss state can be reached by moving the counter at $Z'$ to a new location $Z''$. But then the counter at $Z$ could have been moved to $Z''$ in the first place, so the original configuration was a Win as well.

For every nonzero integer $x$, denote by $v_2(x)$ the exponent of $2$ in the prime decomposition of $x$. Now, given a configuration in which the triangle formed by the three counters is isosceles, the arcs between the vertices having lengths $a, a, b$ respectively (in appropriate units so that $2a + b = n$), we show that the configuration is a Win if and only if $a \neq b$ and $v_2(a-b)$ is odd.

Write $b = a \pm |a-b|$ and notice that the only other isosceles triangle that can be reached from the original configuration is one with arc lengths $a, a \pm |a-b|/2, a \pm |a-b|/2$. If $|a-b|$ is odd, this is of course impossible, so the configuration is a Loss, since all non-isosceles configurations are Wins, by the lemma.

If instead $|a-b|$ is even, then all states that can be reached from the original configuration are Wins, except possibly the state with arc lengths $a, a \pm |a-b|/2, a \pm |a-b|/2$. Consequently, $(a, a, b)$ is a Win if and only if $(a, a \pm |a-b|/2, a \pm |a-b|/2)$ is a Loss. Since the side lengths of this new triangle differ by $|a-b|/2$, the conclusion follows inductively once the exceptional and trivial case $a = b$ is dealt with.

As an immediate corollary, the configuration with arc lengths $1, 1, n-2$ (the starting configuration of the question) is a Win if and only if $v_2(n-3)$ is odd.