Solution:

Initially, all players are in a cycle. Note that once a player leaves the cycle, they cannot rejoin. Furthermore, a new cycle cannot be created. Hence, at any point in time, the graph corresponding to the votes will be a functional graph with a single cycle.

We will first prove that after $\lfloor \log_{2} n \rfloor$ rounds, the cycle will become a self-loop. Then, we will show that in the next $\lfloor \log_{2} n \rfloor$ rounds, all other players vote for the player in the self-loop.

To show the first step, assume the cycle has size $K > 1$ at the beginning of a round. Consider arbitrary player $a$ in the cycle who is updating their vote. Say $a \to b \to c$, all in the cycle. Then $a \to c$ now, bumping $b$ out of the cycle and reducing its size to $K - 1$. Note that $b$ can now update their vote as well without affecting the size of the cycle. If we consider all $K$ original players in the cycle, we see that at least $\lceil \frac{K}{2} \rceil$ of them must still be in the cycle at the time of their update, and hence the cycle's size is reduced to at most $\lfloor \frac{K}{2} \rfloor$. After $\lfloor \log_{2} n \rfloor$ rounds, the cycle must be reduced to size 1.

Now that the cycle has been reduced to a single player, say $z$, consider any path from a player $a$ to $z$. No players can be added to this path now. With a similar argument as the cycle, the length of the path must halve each round. In particular, a path of length $L$ to the cycle gets reduced to length $\lceil \frac{L}{2} \rceil$ (note the ceiling, we had the floor for the cycle). After $\lfloor \log_{2} n \rfloor$ rounds, the path must be reduced to length 1.

Thus, after $\lfloor \log_{2} n \rfloor + \lfloor \log_{2} n \rfloor$ rounds, the graph has been completely reduced. For $n \geq 5$, $\lfloor \log_{2} n \rfloor + \lfloor \log_{2} n \rfloor \leq 2 \lfloor \log_{2} n \rfloor + 1 \leq n$. For the other $n$, we can manually check that $\lfloor \log_{2} n \rfloor + \lceil \log_{2} n \rceil \leq n$.

We will use induction on $n$.

Inductive Hypothesis. Let $G$ be any functional graph with $n$ nodes and a single cycle. Then after $n$ rounds of the given operation, $G$ will become a self-loop with $n - 1$ nodes pointing to it.

Base Case. The cases $n \leq 2$ are clear.

Inductive Step. Assume that the hypothesis is proved for $n = k - 1$ and $n = k - 2$. We will prove it for $n = k$. Consider any initial functional graph with $k$ nodes and a single cycle. Note there is some node $a$ which has in-degree 0 (i.e. no nodes point to it), or all $k$ nodes are in the cycle.

In the first case, consider $G \setminus \{a\}$. Note that all operations except $a$'s own updates are independent of where $a$ is. By the inductive hypothesis, after $k - 1$ rounds, $G \setminus \{a\}$ has become a single self-loop and $k - 2$ nodes pointing to it. Regardless of where $a$ is, it will point to the self-loop after one more round and we are done.

In the case where all $k$ nodes are in a cycle, consider the very first operation $z \to a \to b \implies z \to b, a \to b$. This creates a zero in-degree node $a$, but $z$'s operation has been used for the first round so the inductive hypothesis cannot be naively applied. Instead, consider $b \to c$ (possibly $c = z$ if $k = 3$). At some point in the first round, $b$ will be updated. Either $c$ will become another zero in-degree node, or $a$ will be the only node that points to $c$. Either way, consider $G \setminus \{a, c\}$. By the induction hypothesis, after rounds 2 through $k - 1$, this graph will become a self-loop with $k - 3$ nodes pointing to it. It's also easy to see that $a$ and $c$ both have in-degree 0 after round 2. Then in one more round after round $k - 1$, we must have $a$ and $c$ pointing to the self-loop. So we are done for $n = k$.

By induction, we are done for all $n$.