Convex lattice polygons with equal area and perimeter A convex polygon all of whose vertices have integer coordinates is a convex lattice polygon.

*

*Do there exist mutually non-congruent convex lattice polygons which have the same area and same perimeter?


*If answer to 1 is yes, are there convex lattice polygons which can be cut into some integer number of convex lattice polygons which are not all congruent and all have same area and same perimeter?
Note: The questions have natural analogs in higher dimensions.
 A: This answers question 2 as well. But I think both questions are way more suitable for math.stackexchange.

I add another example because, unlike the first, it has the property that multiplied by $I^{n-2}$ it gives further examples in $\mathbb{R}^n$ as well:

A: My earlier comment "parallelogram and kite" was signaling an infinite family of examples of groups of $m$ convex lattice polygons where all of them from the same group have the same diameter, perimeter, and area (where finite $m$ can be arbitrarily large). There are even two such infinite families.
The FIRST two power families:
Let $\ d_1>d_2>\ldots>d_n>0,\,\ $ and $\,\ y_0 > 0\,\ $ and
$\,\ y_k=y_{k-1}+d_k\ $ for $\ k=1\ldots n.$
Let $\ 0\le a_0<a_1<\ldots a_n\,\ $ and $\,\ x_0:=X\ \ge
\ a_0+ 2\cdot(a_0+\ldots +a_{n-1})+ a_n.$
We start with $4$ vertices $\ (0\ 0)\ $ and $\ (X\ 0)\,\ $ and
$\ (\xi_0\ y_0):=(0\ y_0)\ $ and $\ (\eta_0\ y_0):=(x_0\ y_0).$
Next, we have $\ (\xi_k\ y_k)\,$ and $\ (\eta_k\ y_k),\ $ where either
$$ \xi_k:=\xi_{k-1}+a_k\qquad\text{and}\qquad \eta_k:=\eta_{k-1}-a_{k-1} $$
or
$$ \xi_k:=\xi_{k-1}+a_{k-1}\qquad\text{and}\qquad \eta_k:=\eta_{k-1}-a_k $$
for every $\ k=1\ldots n.$
This way, for fixed $d_k$ and $a_k$ and $X$, we get a family of $2^n$ polygonal when considering all above vertices, and we get an additional $2^n$ polygons when we omit $\ (0\ 0)\ $ and $\ (X\ 0).$ Within each of these two groups the perimeter and the area is respectively the same.
Furthermore, plenty of them can have the same diameter as well. In particular, when you put some additional constraints then all of them within the same group will have the same diameter equal to $\ \sqrt{h_0^{\,2}+X^2}\ $ in the case of $2^{n+1}$-group or $\ X\ $ in the case of the $2^n$-group respectively.
The SECOND two power families:
You take two members (the two can be the same) of one of these families, you take a mirror reflection of one of these two members w.r. to the x-axis in the case of the $2^{n+1}$-group, and w.r. to the line parallel line $y=y_0$. The union set of the vertices of these two polygons forms the vertex set of a respective convex lattice polygon of the SECOND type, and we have a similar situation as in the FIRST case.
A:     
Pick's theorem says these two convex lattice polygons have area
$$i+\frac{b}{2}-1 = 4 + 10/2 -1 = 8 \;,$$
and they both have perimeter $8 + 2 \sqrt{2}$.
You can see I've "bumped out" two corners of an underlying octagon.
(I am interpreting the OP's phrase "the same area and perimeter"
as "the same area and the same perimeter" as opposed to
"the same area and the same perimeter, with the area
numerically equal to the perimeter.")
