Suppose you have $n$ unit squares. Can you dissect each square into polygons such that all the polygons are identical, and then re-arrange the polygons into a single big square of area $n$? Rotations, reflections, translations are allowed, but there can be no gaps or overlaps. All of the polygons across each of the $n$ squares must be identical.
So for example, if $n$ is a square $n=r^2$, the polygons can simply be the squares themselves which you can arrange in an $r\times r$ grid. For a slightly non-trivial example, for $n=2$, you can cut the two squares along a diagonal to get 4 isosceles right triangles each of area $1/2$, and then arrange these into a square of area $2$ by having each edge of the square be a hypotenuse of one of the triangles.
Below I generalize this to integers $n$ that can be written as the sum of two squares. Using the Sum of two squares theorem we can characterize such $n$. But what about other $n$? In particular, is there a solution for $n=3$? Or are there some $n$ for which it is impossible to perform such an operation?
Solution for $n=a^2+b^2$: Suppose $a,b$ are both non-zero; otherwise $n$ itself is a square in which case we are already done. We can then consider the right triangle of sides $a,b,\sqrt{n}$ where $\sqrt{n}$ is the length of the hypotenuse; call the edges of this triangle $A,B,C$, respectively. If we had $4$ copies of this triangle, we can arrange them inside a square of area $n$ (called the "big square") by putting the $C$ edge of each copy as one of the edges of the square. The result is that there will be a square hole of side-length $|b-a|$ in the middle of the big square (rotated relative to the big square of area $n$). Since the inside square had integer side lengths, it can be made of $(b-a)^2$ of our unit squares. So for this inside square, it doesn't matter how we divide the unit squares, since for $(b-a)^2$ of the unit squares we can just re-assemble them back into the original unit squares and then build our inside square of side-length $|b-a|$.
So it remains to divide up the unit squares into identical polygons such that we can build our triangle of side-length $a,b,\sqrt{n}$. This triangle be built from $(ab)^2$ triangles of side-length $\frac{1}{b},\frac{1}{a},\frac{\sqrt{n}}{ab}$. To obtain such triangles, we divide all of our unit squares by first cutting them into $ab$ rectangles of side-lengths $a$ and $b$, and then cutting each rectangle along a diagonal into two right triangles.
You can do the calculation to see that we'll have exactly enough triangles to build the big square after cutting up the unit squares. Or you can just observe that the area of the big square equals the sum of the areas of the original unit squares, so we must have exactly the right number of triangles.