= Solution
Put $\delta=\pi/(2n)$. Between successive measurement bases, the squared overlap is $c=\cos^2\delta$ for retaining the same label and $s=\sin^2\delta$ for changing label. The first step obeys the same rule if we regard the initial state as label 0 at angle zero. Thus the recorded labels form a <Markov chain> with <transition matrix> $\bigl(\begin{smallmatrix}c&s\\s&c\end{smallmatrix}\bigr)$.
Let $q_k$ be the <probability> of label 0 after step $k$. Then $q_0=1$ and
$$
q_k=cq_{k-1}+s(1-q_{k-1}),\qquad 2q_k-1=(c-s)(2q_{k-1}-1).
$$
Since $c-s=\cos(\pi/n)$, iteration yields
$$
\boxed{q_n=\frac12\left[1+\cos^n\left(\frac\pi n\right)\right].}
$$
This is the total final-outcome <probability>, including histories that change label an <even number> of times. The smaller quantity $\cos^{2n}(\pi/(2n))$ counts only the history with label 0 at every step and is not the requested total. For example, at $n=3$ the total is $9/16$, whereas the uninterrupted history has <probability> $27/64$. At $n=1$, the final basis is $\{|1\rangle,|0\rangle\}$ up to signs, so the total <probability> of label 0 is zero, as the formula also gives. This distinguishes nonselective <measurement-guided Zeno transfer> from <postselection> on one record.
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