If $n$ is odd, the required functions are the constant function $f_0 \equiv 2$ along with the $n$ functions $f_i: \{1, \dots, n\} \to \mathbb{N}^*$,
$$f_i(j) = \begin{cases} n+1, & \text{if } j = i, \\ 1, & \text{if } j \neq i, \end{cases} \quad i = 1, \dots, n;$$

notice that $f_0 = f_1$ if $n = 1$. If $n$ is even, $f_0$ is to be removed from the list, by (2). All these functions plainly satisfy the conditions in the statement.
Labelling the positive integers in the list $f(1), \dots, f(n)$ increasingly, $a_1 \le \dots \le a_n$, the problem amounts to determining all lists of positive integers $a_1 \le \dots \le a_n$ such that $(1') \sum_{k=1}^n a_k = 2n$ and $(2') \sum_{k \in K} a_k = n$ for no subset $K$ of the first $n$ positive integers. Notice that $a_n \le n+1$, by $(1')$, and $a_n \ne n$, by $(2')$. With reference again to $(1')$, notice further that if $a_n = n+1$, then the other $a_k$ are all 1, and if $a_n = 2$, then so are the other $a_k$. If $n$ is even, $(2')$ rules out the latter case.
Leaving aside the trivial cases $n=1$ and $n=2$, let $n \ge 3$. To rule out $a_n \ne 2, n+1$, assume, if possible, this is the case, and notice that
$$a_1 - a_n, \quad 0, \quad a_1, \quad a_1 + a_2, \quad \dots, \quad a_1 + a_2 + \dots + a_{n-1}$$
are pairwise distinct integers, so at least two are congruent modulo $n$. Since $a_n \ne 2, n, n+1$, the first two cannot be congruent modulo $n$, and $(2')$ rules out the remaining cases.