Two rectangles: The $1$st has twice the perimeter of the $2$nd and the $2$nd has twice the area of the $1$st.

Let $a,b,c,d$ be as described. Note that this implies

$$a,b,c,d>0$$

else we wouldn't have a rectangle

Now

$$ab=A_1=\frac{1}{2}A_2=\frac{1}{2}cd$$

$$2a+2b=P_1=2P_2=4c+4d$$

This gives us the set of equations

$$2ab=cd$$

$$a+b=2(c+d)$$

Now, this has an infinite number of solutions as we have four unknowns and two equations. Solving for $a$ and $c$ gives us

$$a=\frac{d (b-2 d)}{4 b-d}$$

$$c=\frac{2 b (b-2 d)}{4 b-d}$$

Since these are positive, we know

$$b>2d$$

$$d< 4b$$

which implies $b>2d$. However, we can also write this as

$$b<2d$$

$$d>4b$$

which implies $d>4b$. Thus, one infinite family of solutions might be $b=3d$ which gives us

$$a=\frac{d}{11}$$

$$b=3d$$

$$c=\frac{6d}{11}$$

$$d=d$$

Then for any $d>0$ we get

$$A_1=ab=\frac{6d^2}{22}$$

$$A_2=cd=\frac{6d^2}{11}$$

$$P_1=2a+2b=\frac{68 d}{11}$$

$$P_2=2c+2d=\frac{34 d}{11}$$

which satisfy the original problem. For example, $a=1, b=33, c=6,$ and $d=11$ would give areas of $33$ and $66$ and perimeters of $68$ and $34$, respectively.

Given $a,b,c,d>0$, the condition is $$\begin{cases}c+d=\frac{a+b}2\\ cd=2ab\end{cases}$$

Let's adopt the convention $c\le d$, so that the system is satisfied if and only if $$\begin{cases}c=\frac{a+b}4-\frac12\sqrt{\left(\frac{a+b}2\right)^2-8ab}\\ d=\frac{a+b}4+\frac12\sqrt{\left(\frac{a+b}2\right)^2-8ab}\end{cases}\\ \begin{cases}c=\frac{a+b-\sqrt{a^2-30ab+b^2}}4\\ d=\frac{a+b+\sqrt{a^2-30ab+b^2}}4\end{cases}$$

Notice that, under the hypothesis, both those numbers are strictly positive, provided that the square root exists, i.e. provided that $a^2-30ab+b^2\ge 0$. Under the condition $a,b>0$ this is the case if and only if $\frac ab\le 15-4\sqrt{14}$ or $\frac ab\ge15+4\sqrt{14}$.

Therefore the set of solutions $(c,d)$ with $c\le d$ is parametrised by the region of plane $\begin{cases} \frac ab\ge 15+4\sqrt{14}\\ b>0 \end{cases}\lor\begin{cases} \frac ab\le 15-4\sqrt{14}\\ b>0 \end{cases}$ via the function $$(a,b)\mapsto \left(\frac{a+b-\sqrt{a^2-30ab+b^2}}4,\frac{a+b+\sqrt{a^2-30ab+b^2}}4\right)$$