In both solutions, by a polygon we always mean its interior together with its boundary. We start with finding a regular $n$-gon $\mathcal{C}$ which (i) is inscribed into $\mathcal{B}$ (that is, all vertices of $\mathcal{C}$ lie on the perimeter of $\mathcal{B}$); and (ii) is either a translation of $\mathcal{A}$, or a homothetic image of $\mathcal{A}$ with a positive factor. Such a polygon may be constructed as follows. Let $O_{A}$ and $O_{B}$ be the centers of $\mathcal{A}$ and $\mathcal{B}$, respectively, and let $A$ be an arbitrary vertex of $\mathcal{A}$. Let $\overrightarrow{O_{B} C}$ be the vector co-directional to $\overrightarrow{O_{A} A}$, with $C$ lying on the perimeter of $\mathcal{B}$. The rotations of $C$ around $O_{B}$ by multiples of $2 \pi / n$ form the required polygon. Indeed, it is regular, inscribed into $\mathcal{B}$ (due to the rotational symmetry of $\mathcal{B}$), and finally the translation/homothety mapping $\overrightarrow{O_{A} A}$ to $\overrightarrow{O_{B} C}$ maps $\mathcal{A}$ to $\mathcal{C}$. Now we separate two cases. ! Construction of $\mathcal{C}$ ! Case 1: Translation

Case 1: $\mathcal{C}$ is a translation of $\mathcal{A}$ by a vector $\vec{v}$. Denote by $t$ the translation transform by vector $\vec{v}$. We need to prove that the vertices of $\mathcal{C}$ which stay in $\mathcal{B}$ under $t$ are consecutive. To visualize the argument, we refer the plane to Cartesian coordinates so that the $x$-axis is co-directional with $\vec{v}$. This way, the notions of right/left and top/bottom are also introduced, according to the $x$- and $y$-coordinates, respectively.

Let $B_{\mathrm{T}}$ and $B_{\mathrm{B}}$ be the top and the bottom vertices of $\mathcal{B}$ (if several vertices are extremal, we take the rightmost of them). They split the perimeter of $\mathcal{B}$ into the right part $\mathcal{B}_{\mathrm{R}}$ and the left part $\mathcal{B}_{\mathrm{L}}$ (the vertices $B_{\mathrm{T}}$ and $B_{\mathrm{B}}$ are assumed to lie in both parts); each part forms a connected subset of the perimeter of $\mathcal{B}$. So the vertices of $\mathcal{C}$ are also split into two parts $\mathcal{C}_{\mathrm{L}} \subset \mathcal{B}_{\mathrm{L}}$ and $\mathcal{C}_{\mathrm{R}} \subset \mathcal{B}_{\mathrm{R}}$, each of which consists of consecutive vertices.

Now, all the points in $\mathcal{B}_{\mathrm{R}}$ (and hence in $\mathcal{C}_{\mathrm{R}}$) move out from $\mathcal{B}$ under $t$, since they are the rightmost points of $\mathcal{B}$ on the corresponding horizontal lines. It remains to prove that the vertices of $\mathcal{C}_{\mathrm{L}}$ which stay in $\mathcal{B}$ under $t$ are consecutive.

For this purpose, let $C_{1}, C_{2}$, and $C_{3}$ be three vertices in $\mathcal{C}_{\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $t\left(C_{1}\right)$ and $t\left(C_{3}\right)$ lie in $\mathcal{B}$; we need to prove that $t\left(C_{2}\right) \in \mathcal{B}$ as well. Let $A_{i}=t\left(C_{i}\right)$. The line through $C_{2}$ parallel to $\vec{v}$ crosses the segment $C_{1} C_{3}$ to the right of $C_{2}$; this means that this line crosses $A_{1} A_{3}$ to the right of $A_{2}$, so $A_{2}$ lies inside the triangle $A_{1} C_{2} A_{3}$ which is contained in $\mathcal{B}$. This yields the desired result.

Case 2: $\mathcal{C}$ is a homothetic image of $\mathcal{A}$ centered at $X$ with factor $k>0$.

Denote by $h$ the homothety mapping $\mathcal{C}$ to $\mathcal{A}$. We need now to prove that the vertices of $\mathcal{C}$ which stay in $\mathcal{B}$ after applying $h$ are consecutive. If $X \in \mathcal{B}$, the claim is easy. Indeed, if $k>1$, then the vertices of $\mathcal{A}$ lie on the extensions of such segments $X C$ beyond $C$, and almost all these extensions lie outside $\mathcal{B}$. The exceptions may occur only in case when $X$ lies on the boundary of $\mathcal{B}$, and they may cause one or two vertices of $\mathcal{A}$ to stay on the boundary of $\mathcal{B}$. But even in this case those vertices are still consecutive.

So, from now on we assume that $X \notin \mathcal{B}$. Now, there are two vertices $B_{\mathrm{T}}$ and $\mathcal{B}_{\mathrm{B}}$ of $\mathcal{B}$ such that $\mathcal{B}$ is contained in the angle $\angle B_{\mathrm{T}} X B_{\mathrm{B}}$; if there are several options, say, for $B_{\mathrm{T}}$, then we choose the farthest one from $X$ if $k>1$, and the nearest one if $k<1$.

Subcase 2.1: $k>1$. In this subcase, all points from $\mathcal{B}_{\mathrm{R}}$ (and hence from $\mathcal{C}_{\mathrm{R}}$) move out from $\mathcal{B}$ under $h$, because they are the farthest points of $\mathcal{B}$ on the corresponding rays emanated from $X$. It remains to prove that the vertices of $\mathcal{C}_{\mathrm{L}}$ which stay in $\mathcal{B}$ under $h$ are consecutive.

Again, let $C_{1}, C_{2}, C_{3}$ be three vertices in $\mathcal{C}_{\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $h\left(C_{1}\right)$ and $h\left(C_{3}\right)$ lie in $\mathcal{B}$. Let $A_{i}=h\left(C_{i}\right)$. Then the ray $X C_{2}$ crosses the segment $C_{1} C_{3}$ beyond $C_{2}$, so this ray crosses $A_{1} A_{3}$ beyond $A_{2}$; this implies that $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\mathcal{B}$. !

Subcase 2.2: $k<1$. This case is completely similar to the previous one. All points from $\mathcal{B}_{\mathrm{L}}$ (and hence from $\mathcal{C}_{\mathrm{L}}$) move out from $\mathcal{B}$ under $h$, because they are the nearest points of $\mathcal{B}$ on the corresponding rays emanated from $X$. Assume that $C_{1}, C_{2}$, and $C_{3}$ are three vertices in $\mathcal{C}_{\mathrm{R}}$ such that $C_{2}$ lies between $C_{1}$ and $C_{3}$, and $h\left(C_{1}\right)$ and $h\left(C_{3}\right)$ lie in $\mathcal{B}$; let $A_{i}=h\left(C_{i}\right)$. Then $A_{2}$ lies on the segment $X C_{2}$, and the segments $X A_{2}$ and $A_{1} A_{3}$ cross each other. Thus $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\mathcal{B}$.

Comment 1. In fact, Case 1 can be reduced to Case 2 via the following argument. Assume that $\mathcal{A}$ and $\mathcal{C}$ are congruent. Apply to $\mathcal{A}$ a homothety centered at $O_{B}$ with a factor slightly smaller than 1 to obtain a polygon $\mathcal{A}^{\prime}$. With appropriately chosen factor, the vertices of $\mathcal{A}$ which were outside/inside $\mathcal{B}$ stay outside/inside it, so it suffices to prove our claim for $\mathcal{A}^{\prime}$ instead of $\mathcal{A}$. And now, the polygon $\mathcal{A}^{\prime}$ is a homothetic image of $\mathcal{C}$, so the arguments from Case 2 apply.

Comment 2. After the polygon $\mathcal{C}$ has been found, the rest of the solution uses only the convexity of the polygons, instead of regularity. Thus, it proves a more general statement:

Assume that $\mathcal{A}, \mathcal{B}$, and $\mathcal{C}$ are three convex polygons in the plane such that $\mathcal{C}$ is inscribed into $\mathcal{B}$, and $\mathcal{A}$ can be obtained from it via either translation or positive homothety. Then the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive.