We shall prove that this can only happen whenever $n$ is prime. We shall prove the following lemma;

Lemma 1. Let $p$ be a prime number and $x_1, \cdots, x_r$ be pair-wise distinct (mod $p$) and $y_1, \cdots, y_r$ be distinct positive integers. Then, there is a polynomial $P(x)$ with integer coefficients such that
$$P(x_i) \equiv y_i \pmod{p}$$
Proof. Adopting the Lagrange Interpolation Formula and constructing the following polynomial
$$\sum a_i \cdot \frac{(x-x_1)\cdots(x-x_r)}{(x-x_i)}$$
where $a_i = y_i/((x_i-x_1)\cdots(x_i-x_{i-1})(x_i-x_{i+1})\cdots(x_i-x_n))$, where by fractions, we mean the multiplicative inverses modulo $p$. It is clear that this polynomial satisfies the condition of the lemma. This completes our proof.

Notice also that if $n$ is an arbitrary integer and $\gcd(x_i - x_j, n) = 1$ the preceding lemma remains true. Hence, if $A$ and $B$ are two arbitrary sets with pair-wise distinct elements mod $p$ such that $|A| \ge |B|$, using the above lemma, we find that $A$ would be translatable to $B$.
Let $n > 4$ be a composite number and $p$ be its smallest prime divisor, it follows that $n > 3p$. Now, consider $A = \{1, p+1, 2p+1\}$ and $B = \{1, 2\}$. We prove that $B$ would not be translatable to $A$. On contrary, we shall have $P(R) \equiv 1, P(S) \equiv 2 \pmod{p}$, for some $R, S \in A$. On the other hand, $R \equiv S \pmod{p}$ hence, $P(R) \equiv P(S) \pmod{p}$, contradiction. ■