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tiles ${\bf Z}^d$ for some $d$: Vytautas Gruslys, Imre Leader, and Ta Sheng Tan: Tiling with arbitrary tiles. Proc. London Math. Soc. (2016) 112 (6): 1019-1039. …

There are two questions here: Q1) Which convex pentagons tile the plane? Q2) What are all tilings of the plane by copies of a single convex pentagon? The Wikipedia page you cite concerns Question 1 …

Google soon finds that Q2 is problem C11 in Unsolved Problems in Geometry by Croft, Falconer, and Guy. But perhaps it's been solved during the intervening decades. URL

T_1$ such that $T_1 - T_0$ is the union of three acute, non-isosceles triangles with circumradii distinct from each other and from $s$; likewise inscribe $T_1$ into $T_2$, and $T_2$ into $T_3$, etc., tiling … Now connect each of these triangles' vertices to its circumcenter to obtain a tiling of the plane by isosceles triangles any two of which have distinct repeated sides, and thus a fortiori are not congruent …

There are two choices at each stage, most of which yield a tiling of the plane; alternatively, choose the $T'_n$ and $T_n$ to all share a vertex with $T_0$, getting a geometric progression [sic] of triangles …

This picture shows how $R$ is also the hexagonal tiling by copies of $H$ with every third hexagon removed. Hence three parallel copies of $R$ constitute a perfect $2$-covering of the plane. … H}-H$ and $R$, so we have our family of convex pentagons with a minimal tiling thickness of $2$. P.S. …

This picture looks like a counterexample with $N=2$ and $R$ a convex pentagon: This should work more generally starting from an $n \times (n+1)$ rectangles for any integer $n>1$, removing two congrue …