Problem 1863

The answer is $p^n$.
Let $a_i = \frac{(1+p)^i - 1}{p}$ for $i \ge 0$. By induction we obtain $f(a_i) \equiv i f(1) \pmod{p^n}$ for all $i \ge 0$.
If $p > 2$ or $p = 2$ and $n = 1$, then
$$a_i \equiv a_j \pmod{p^n} \Rightarrow (1+p)^{j-i} \equiv 1 \pmod{p^{n+1}} \Rightarrow i \equiv j \pmod{p^n}.$$
Since $a_i + a_j + p a_i a_j = a_{i+j}$ for all $i, j \ge 0$, the function defined by $f(a_i) \equiv i c \pmod{p^n}$, $(0 \le i < p^n)$, satisfies the condition of the question for any choice of $c \in \mathbb{Z}_{p^n}$.

$$a_i \equiv a_j \pmod{2^n} \Rightarrow 3^{j-i} \equiv 1 \pmod{2^{n+1}} \Rightarrow i \equiv j \pmod{2^{n-1}}$$
and $Z_{p^n} = \{a_i : 0 \le i < 2^{n-1}\} \cup \{-a_i - 1 : 0 \le i < 2^{n-1}\}$. We also have $2f(-1) \equiv f(-1) + f(-1) \equiv f((-1) + (-1) + 2(-1)(-1)) \equiv f(0) \equiv 0 \pmod{2^n}$.
Since $a_i - a_j + 2 a_i a_j = a_{i+j}$, $a_i + (-a_j - 1) + 2 a_i (-a_j - 1) = -a_{i+j} - 1$, and $(-a_i - 1) + (-a_j - 1) + 2(-a_i - 1)(-a_j - 1) = a_{i+j}$ for all $i, j \ge 0$, the function defined by $f(a_i) \equiv i c \pmod{2^n}$ and $f(-a_i - 1) \equiv i c + d \pmod{2^n}$, ($0 \le i < 2^{n-1}$), satisfies the condition of the question for any choice of $c \in 2\mathbb{Z}_{2^n}$ and $d \in 2^{n-1}\mathbb{Z}_{2^n}$ as $f(a) + f(-1) \equiv f(-a - 1) \pmod{2^n}$ for all $a \in Z_{p^n}$.