Given an arbitrary code-set for the lock, if two adjacent vertices have different numbers, we label the sides linking them by letter $a$; if they have different colors, we label it by $b$; if both the numbers and colors are the same, we label it by $c$. Once the number and color on vertex $A_1$ are set (there are four different sets for it), we can then set $A_2, A_3, \dots, A_n$ one by one according to the letters labelled on each side. In order to let it return to the initial set of $A_1$ finally, the numbers of sides labelled $a$ and $b$ must be both even. So the number of code-sets for the lock is four times the number of labelled-side sequences which satisfy the condition that the numbers of sides labelled by $a$ and $b$ are both even.

Suppose there are $2i$ ($0 \le i \le \lfloor \frac{n}{2} \rfloor$) sides labelled by $a$, and $2j$ ($0 \le j \le \lfloor \frac{n-2i}{2} \rfloor$) sides labelled by $b$. Then there are $C_n^{2i}$ ways to label $2i$ sides by $a$ from $n$ ones, $C_{n-2i}^{2j}$ ways to label $2j$ sides by $b$ from $n-2i$ ones, and the remaining sides are labelled by $c$. Therefore, by the Multiplication Principle, there are $C_n^{2i} C_{n-2i}^{2j}$ ways to label all the sides. So there are totally
$$4 \sum_{i=0}^{\lfloor \frac{n}{2} \rfloor} \left( C_n^{2i} \sum_{j=0}^{\lfloor \frac{n-2i}{2} \rfloor} C_{n-2i}^{2j} \right)$$
code-sets for the lock. Here we stipulate $C_0^0 = 1$.

When $n$ is odd, we have $n - 2i > 0$, and then
$$\sum_{j=0}^{\left\lfloor \frac{n-2i}{2} \right\rfloor} C_{n-2i}^{2j} = 2^{n-2i-1}.$$
Substituting it into the previous formula, we get
$$\begin{aligned} 4 \sum_{i=0}^{\left\lfloor \frac{n}{2} \right\rfloor} \left( C_n^{2i} \sum_{j=0}^{\left\lfloor \frac{n-2i}{2} \right\rfloor} C_{n-2i}^{2j} \right) &= 4 \sum_{i=0}^{\left\lfloor \frac{n}{2} \right\rfloor} (C_n^{2i} 2^{n-2i-1}) \\ &= 2 \sum_{i=0}^{\left\lfloor \frac{n}{2} \right\rfloor} (C_n^{2i} 2^{n-2i}) \\ &= \sum_{k=0}^{n} C_n^k 2^{n-k} + \sum_{k=0}^{n} C_n^k 2^{n-k} (-1)^k \\ &= (2+1)^n + (2-1)^n \\ &= 3^n + 1. \end{aligned}$$

When $n$ is even, if $i < \frac{n}{2}$, then the above remains true; if $i = \frac{n}{2}$, then all the sides of the polygon are labelled by $a$, and that means there is only one way to label the sides. Therefore, there are totally
$$\begin{aligned} 4 \sum_{i=0}^{\left\lfloor \frac{n}{2} \right\rfloor} \left( C_n^{2i} \sum_{j=0}^{\left\lfloor \frac{n-2i}{2} \right\rfloor} C_{n-2i}^{2j} \right) &= 4 \times \left( 1 + \sum_{i=0}^{\left\lfloor \frac{n}{2} \right\rfloor} (C_n^{2i} 2^{n-2i-1}) \right) \\ &= 2 + 4 \sum_{i=0}^{\left\lfloor \frac{n}{2} \right\rfloor} (C_n^{2i} 2^{n-2i-1}) \\ &= 3^n + 3 \end{aligned}$$
code-sets for the lock.

In summary,
the number of code-sets for the lock is
$$\begin{cases} 3^n + 1 & \text{when } n \text{ is odd,} \\ 3^n + 3 & \text{when } n \text{ is even.} \end{cases}$$