Permutation of Disjoint Sets of a Symmetric Group

Question

Problem Description:

Consider a symmetric group $S_n$ acting on $n$ objects. We partition $S_n$ into two sets $A, B$ such that

1. $A \cap B= \emptyset$ and $A \cup B = S_n$. In other words, $S_n$ is disjoint union of $A$ and $B$.

2. There exists a permutation $\pi \in S_n$ such that $A \circ {\pi} =B$. In other words, every product $a \circ{\pi}$ (where $a \in A$) is an element $b$ of $B$, and conversely, every $b \in B$ can be written in this way.

Question:

For which $n$ we can have $S_n$ that can be partitioned above way?

Edit:

Construction of product can be $\pi \circ A =B$ instead of $A \circ {\pi} =B$. For computational convenience , consider $A \circ {\pi} =B$.

Answer 1

Any $n \geq 2$. Partition according to parity so that $A = A_n$ and $B = S_n - A_n$. Take $\pi = (1, 2)$, which is an odd permutation. Since any even permutation in $A$ multiplied by the odd permutation $\pi$ will result in an odd permutation in $B$, we are done.