Denote the sum of labels of $A$ and $B$ by $x$ and $y$, respectively. Then $y = 2x$. Thus, we have
$$3x = x + y = 1 + 2 + \cdots + 12 = 78, \quad x = 26.$$
Obviously, $1, 2 \in A$ and $11, 12 \in B$. Denote $A = \{1, 2, a, b, c, d\}$, where $a < b < c < d$. Then $a + b + c + d = 23$, and $a \ge 3$, $8 \le d \le 10$ (if $d \le 7$, then $a + b + c + d \le 4 + 5 + 6 + 7 = 22$, which is a contradiction.)

(1) If $d = 8$, then $A = \{1, 2, a, b, c, 8\}$, $c \le 7$, $a + b + c = 15$. Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, that is, $A = \{1, 2, 3, 5, 7, 8\}$ or $A = \{1, 2, 4, 5, 6, 8\}$.

If $A = \{1, 2, 3, 5, 7, 8\}$, then $B = \{4, 6, 9, 10, 11, 12\}$. Since $B$ contains $11, 4, 6$ and $12$, there is only one tower that, in $A$, $8$ and $3$, $3$ and $1$, $1$ and $5$, $5$ and $7$ are adjacent.

If $A = \{1, 2, 4, 5, 6, 8\}$, then $B = \{3, 7, 9, 10, 11, 12\}$. Similarly, we see that, in $A$, $1$ and $2$, $5$ and $6$, $4$ and $8$ are adjacent, respectively. There are two arrangements, that is, there are two towers.

(2) If $d = 9$, then $A = \{1, 2, a, b, c, 9\}$, $c \le 8$, $a + b + c = 14$, where $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, that is, $A = \{1, 2, 3, 5, 6, 9\}$ or $A = \{1, 2, 3, 4, 7, 9\}$.

If $A = \{1, 2, 3, 5, 6, 9\}$, then $B = \{4, 7, 8, 10, 11, 12\}$. To obtain $4$, $10$ and $12$ in $B$, $1$, $3$, and $9$ in $A$ must be adjacent pairwise, it is impossible!

If $A = \{1, 2, 3, 4, 7, 9\}$, then $B = \{5, 6, 8, 10, 11, 12\}$. To obtain $6$, $8$ and $12$ in $B$, $2$ and $4$, $1$ and $7$, $9$ and $3$ must be adjacent in $A$, respectively. There are two arrangements, that is, there are two towers.

(3) If $d = 10$, then $A = \{1, 2, a, b, c, 10\}$, where $c \le 9$, $a + b + c = 13$. Thus, $(a, b, c) = (3, 4, 6)$, that is, $A = \{1, 2, 3, 4, 6, 10\}$ and $B = \{5, 7, 8, 9, 11, 12\}$. To obtain $8$, $9$, $11$ and $12$ in $B$, $6$ and $2$, $6$ and $3$, $10$ and $1$, $10$ and $2$ must be adjacent, respectively. There is only one tower.

Summing up, there are six different towers all together. ☐