Since $f^{2005} = \textrm{id}$, it follows $f$ is a permutation, hence a product of cycles $\prod_{k=0}^s \gamma_k$, of lengths $\lambda_k$, where $\gamma_0 = (1)$, so $\lambda_0 = 1$ (because we are given $1$ is a fixed point). We must therefore have $\sum_{k=1}^s \lambda_k = 1599$, and $\lambda_k \mid 2500 = 5\cdot 401$ for all $1\leq k \leq s$. But then it means $\lambda_k \in \{1,5,401\}$. Since $1599 = 5\cdot 401 - 401 - 5$, it is impossible to have all $\lambda_k \neq 1$, so there is (at least) one more fixed point. For any $n>1600$ one can express $n-1$ as a sum with terms just $5$ and $401$ (Sylvester problem, or Frobenius, or the dreadfully-named Chicken McNugget theorem), so the answer is NO, no need to have that extra fixed point.