Two equal polygons $F$ and $F'$ are given on the plane. It is known that the vertices of the polygon $F$ belong to $F'$ (may lie inside it or on the border). Is it true that all the vertices of these polygons coincide?
1. **Define the polygons and their properties:**
   - Let $F$ and $F'$ be two equal polygons on the plane.
   - The vertices of polygon $F$ belong to polygon $F'$, meaning each vertex of $F$ is either on the boundary or inside $F'$.

2. **Construct a counterexample:**
   - Consider an equilateral triangle $ABC$.
   - Let $M$ and $N$ be points symmetric with respect to the median $AA'$ of triangle $ABC$.
   - Construct polygon $F$ as $ABCM$ and polygon $F'$ as $ACBN$.

3. **Analyze the properties of $F$ and $F'$:**
   - Both $F$ and $F'$ are quadrilaterals with vertices $A$, $B$, $C$, and an additional point ($M$ for $F$ and $N$ for $F'$).
   - Since $M$ and $N$ are symmetric with respect to the median $AA'$, the quadrilaterals $F$ and $F'$ are congruent (equal in shape and size).

4. **Verify the vertices' positions:**
   - The vertices of $F$ are $A$, $B$, $C$, and $M$.
   - The vertices of $F'$ are $A$, $C$, $B$, and $N$.
   - Since $M$ and $N$ are symmetric with respect to the median $AA'$, $M$ does not coincide with $N$.

5. **Conclusion:**
   - Even though $F$ and $F'$ are equal polygons and the vertices of $F$ belong to $F'$, not all vertices of $F$ and $F'$ coincide.
   - This provides a counterexample to the statement that all vertices of these polygons must coincide.

\[
\boxed{\text{False}}
\]