There are $n$ coins in a row. Two players take turns picking a coin and flipping it. The location of the heads and tails should not repeat. Loses the one who can not make a move. Which of player can always win, no matter how his opponent plays?
1. Let's analyze the problem by considering the possible moves and the constraints. Each player takes turns flipping a coin, and the sequence of heads and tails should not repeat. The player who cannot make a move loses.

2. To determine the winning strategy, we need to consider the parity (odd or even) of the number of coins, \( n \).

3. If \( n \) is odd:
   - The first player can always flip the same coin. Since the number of coins is odd, the first player will always have a move to make, while the second player will eventually run out of moves. This is because the first player can always flip the coin back to its original state, ensuring that the sequence of heads and tails does not repeat.

4. If \( n \) is even:
   - The situation is more complex. The first player cannot simply flip the same coin repeatedly, as the second player will eventually force a situation where the first player has no valid move. In this case, the second player can always respond to the first player's move by flipping a different coin, ensuring that the sequence of heads and tails does not repeat.

5. Therefore, the first player can always win if \( n \) is odd, and the second player can always win if \( n \) is even.

Conclusion:
The first player can always win if the number of coins \( n \) is odd. The second player can always win if the number of coins \( n \) is even.

The final answer is: The first player can always win if \( \boxed{ n } \) is odd, and the second player can always win if \( n \) is even.