5. Rotate a regular $n$-sided polygon $M$ around its center by an angle of $\frac{\pi}{n}$ to get $M^{\prime}$. How many convex $n$-sided polygons can the figure $M \cup M^{\prime}$ be divided into at minimum?
5. The minimum number of convex polygons is $n+1$. A division of the figure $M \cup M^{\prime}$ into $n+1$ convex polygons clearly exists; we only need to prove that it cannot be divided into fewer convex polygons. For this, we examine all the "outer" vertices of the figure $M \cup M^{\prime}$ and number the convex polygons. Since we want to minimize the number of convex polygons, each "outer" vertex belongs to only one convex polygon. Suppose the "outer" vertices belonging to the $i$-th polygon are colored the $i$-th color. Thus, the problem reduces to the following lemma:

Lemma: On a circle, there are $m$ points, each colored with one of $k$ colors (each point is colored with one of these colors), such that no two adjacent points have the same color, and if points $A$ and $B$ are colored the same, and points $C$ and $D$ are colored the same but with a different color, then segments $AB$ and $CD$ do not intersect. Then the number of colors used $k_{m} \geqslant\left[\frac{m+3}{2}\right]$, where $m \geqslant 2$.

This lemma can be proven by induction on $m$.
When $m=2$ and 3, it is clearly true that $k=2$ and 3, respectively, so the claim holds. Assume the lemma's claim holds for up to $m-1$ points, and consider the case of $m$ points.
Let $A$ and $B$ be two points of the same color, then $A$ and $B$ are not adjacent, and $m \geqslant 4$. Draw a line $L$ to separate $A$ and $B$, ensuring that $L$ has points on both sides (as shown in Figure 1), and that $L$ passes between $A$ and its neighbor $Y$ (in the counterclockwise direction), and between $B$ and its neighbor $X$ (also in the counterclockwise direction). Connect $AX$ and $BY$, thus dividing the circle into two smaller "circles" (each on one side of $L$). One of these "circles" has $2 \leqslant m_{1} \leqslant m-2$ points, and the other has $2 \leqslant m - m_{1} \leqslant m-2$ points. Clearly, the induction hypothesis can be applied to each of them. However, since they use the same color (as $A$ and $B$ are the same color), there cannot be any other points of the same color between these two smaller "circles" (otherwise, the line connecting them would intersect $AB$). Therefore, we have
$$
\begin{aligned}
k_{m} & \geqslant k_{m_{1}} + k_{m - m_{1}} - 1 \\
& \geqslant \left[\frac{m_{1} + 3}{2}\right] + \left[\frac{m - m_{1} + 3}{2}\right] - 1 \\
& \geqslant \left[\frac{m + 3}{2}\right],
\end{aligned}
$$

Thus, the lemma's claim holds for all $m \geqslant 2$.
Since one color corresponds to one of the divided polygons, the number of polygons is at least $\left[\frac{2n + 3}{2}\right] = n + 1$.