Suppose that $x, y \in \{1,\dots, n\}$. Is it possible to find a nonzero $f \in \mathbb{C}$ such that
$$f(x+y) = f(x)f(y) \text{ for } x \neq y$$
$$f(x)^2 - \frac{1}{2x-1} = f(2x)$$
This problem is related to solving $f(x+y) = f(x)f(y) $ with solution $a^x$.
There is a trivial solution for $\ n=1\ $ and a simple solution for $\ n=2.\ $ It is however straightforward to see that there does not exist any solution for any $\ n\ge 4.\ $ I'll leave the pleasure of verifying this to the readers. (You may consult my comments above, under the OP's Question).