Let us define a **perfect congruent partition** of a planar region $R$ as a partition of it with no portion left over into some finite number n of pieces that are all mutually congruent (ie any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve).

**Note:** It is known there are convex planar regions - indeed quadrilaterals - which do not allow perfect congruent partition for any n ([1] proves a stronger result).

**Claim:** If a convex polygonal $R$ allows a perfect congruent partition of itself into $N$ non-convex pieces each with finitely many sides, then $R$ also allows a perfect congruent partition into $N$ convex pieces with finitely many sides. In other words, allowing the pieces to be non-convex polygons does not improve the chances of a convex planar region achieving a perfect congruent partition into $N$ pieces.

I know no proof, no counter example. One can consider replacing 'congruent' with 'similar' in the above question. Some more related thoughts are in [2].

References:

1.https://www.research.ibm.com/haifa/ponderthis/challenges/December2003.html 2.https://arxiv.org/abs/1002.0122

**A bit added on March 15th, 2024:** Are there convex polygonal regions with even number of edges that has this property of being perfect congruent partitionable only into 2 non-convex pieces?