I am asked to find if there is an affine isomophism $f: R^2 \rightarrow R^2$ that takes one parallelogram into another one, for any two given parallelograms (in $R^2$).
Taking into account that both translations and rotations are affine isomorphisms from $R^2$ to itself, I do the following:
Let $P_1$ and $P_2$ be the two given parallelograms and $C_i$ be the centre of parallelogram $P_i$. If $T_v$ is a translation with $v = C_2 - C_1$ then $T_v$ takes $P_1$ and centers it with $P_2$. Without loss of generality we now have $T_v(C_1) = C_2 = (0,0)$ i.e. the parallelograms have their centers in the origin. Picking now one diagonal $d_i$ for each parallelogram and by calculating the angle between $d_1$ and $d_2$ I can rotate $P_1$ so that both parallelograms have their center in the origin and have one diagonal aligned.
If now both diagonals are aligned, I can prove that there exists the said isomorphism: we know that $T_\alpha(x, y) = (\alpha x, y)$ and $T_\beta(x, y) = (x, \beta y)$ are linear transformations. Again and without loss of generality we can assume that if both diagonals are aligned, then both diagonals will also be aligned with the $Ox$ and $Oy$ axis (we can take a vector from each diagonal and make them the new basis of our coordinate system). Having both diagonals aligned with the axis, we just have to pick the right $\alpha$ and $\beta$ so that the four vertices of $P_1$ coincide with the vertices of $P_2$.
I struggle only when the second diagonal isn't aligned. That happens exactly when the angle between $P_1$'s diagonals differs from the angle between $P_2$'s diagonals. If that is the case, I can't prove nor disprove the existence of said affine isomorphism, nor do I know how to take it from here. Can anyone pick it from here?
Also, from the context of the question, this should be rather simple. Either because there exists a simple affirmative proof or a simple counter-example. I can't think of any. Feel free to chip in.
Let $P_1$ and $P_2$ be any two parallelograms and let $V_i$ be a vertex of parallelogram $P_i$. We can assume, without loss of generality, that both $V_1$ and $V_2$ are at the origin of my referential.
(If they were not, I could translate each one of them, separately, to the origin, using two translations $T_1$ and $T_2$ that are affine isomorphisms).
Now for each parallelogram we consider the vectors induced by the two sides that share the vertex $V_i$. For each parallelogram, separately, the two vectors form a basis of $\Bbb{R}^2$ because if they did not, they were collinear and we would not have a parallelogram. Now we have two different basis $B_1$ and $B_2$ for $\Bbb{R}^2$ but as we know from trivial linear algebra there is a linear transformation $M$ (and hence an affine isomorphism) that changes from base $B_1$ to base $B_2$ and vice-versa. Therefore we can take any side from $P_1$ (thinking of it as a vector) to the other corresponding vector by the composition of the three affine isomorphisms mentioned: $T_2\ o\ M\ o\ T_1$
Or conversely with its inverse.
Another proof would use the fact that if $A$ is an affine space of dimension $n$, there is an affine isomorphism that takes $n+1$ independent points to any other $n+1$ independent points. We could just take 3 vertices from one parallelogram, the corresponding 3 vertices of the other, create $\phi$ that takes each of those 3 to the corresponding, and show that the fourth vertex is also being taken to its corresponding vertex.