Tiling With Unique Neighborhoods - 2

Here, I try to gather what was written as an addition to this earlier post (https://nandacumar.blogspot.com/2019/03/more-on-2d-layouts-with-unique.html) into a proper self-contained post.

Given any region that can tile the plane (without gaps or overlaps) and a tiling layout with it: define the 'neighborhood' of any given unit tile T in the layout as the union of T and all other tiles in the layout that touch it.

Question: Does there exist a tile (not necessarily convex) that can form a tiling layout such that if we choose any two tiles in the layout, their neighborhoods are non-congruent?

Further, Is this question related to the 'Einstein problem', the question of finding aperiodic tilings with a single tile (An aperiodic tiling is a non-periodic tiling with the additional property that it does not contain arbitrarily large periodic patches)? Probably the two questions are different but I am not sure.... and yes, our question has obvious higher dimensional versions.

Some further thoughts:

It looks easy to a tile the plane with all neighborhoods non-congruent if we use two different species of tiles. Consider any rectangle A (A could be the unit square) and another rectangle B such that ratio between the dimensions of B to those of A are irrational. Form infinite strips of As and Bs and put them side to side thus: (1)keep one side of a unit in each of the strips - whether A-strip or B-strip - flush with the Y axis (all strips are now obviously parallel to X axis). (2) Now, keeping the A-strips fixed, slide each B-strip by a mutually independent small irrational distance in X direction. Such an arrangement looks enough to ensure that every unit rectangle - of type A or B - has a unique neighborhood.

We note here that there seem to be enough small irrationals (due to properties of the various 'alephs') to form an (countably) infinite number of different tilings for the plane - each with the unique neighborhood property - with the same A and B rectangles.
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Claim: With unit cubes, we can form a tiling (a tessellation, to be precise) of 3D space with every unit possessing a unique neighborhood.

Proof: Consider a corner to corner layout of identical squares that tile the plane. Assume the X and Y axes to be aligned along the edges of one of the squares. Call this tiling layout T1.

Now, consider another tiling layout T2 that is identical to T1 and lying directly on top of T1. Keeping T1 fixed, rotate T2 about Z axis by any angle α with its sine and cosine irrational and mutually linearly independent over rationals (there are infinitely many such values of α as per Niven's theorem, say). Now we can consider T2 to be formed by the basis vectors: (cos α, sin α) and (-sin α, cos α). Obviously, vertices on T2 are formed by linear combinations of these vectors with coefficients chosen from integers.
Let us call the fractional part of a number as its 'offset'. We observe that the offsets of the X coordinates (likewise, Y coordinates)of the vertices of T2 are all unique. Indeed, if two vertices on T2 have same X( or Y) coordinate, then, the X ( or Y) component of the vector joining these two vertices, which will be an integer, can be written as a linear combination - with integer coefficients - of sin α and cos α which is impossible due to irrationality. Now, every square in T1 is overlaid by a certain number of vertices and squares from T2. Since every 'T2 square' has unique offsets, we have: for every 'T1 square' S, the portion of T2 it sees has a unique position with respect to S (note that the T2 vertices that lie within S will have unique X and Y offsets).

Note: Observe that there is a 90 degree rotation symmetry to the above construction. This can be broken by translating T2 in a general 2D direction by some small irrational distance thus ensuring every S in T1 seeing a unique portion of T2.

Now, let all squares of layouts T1 and T2 be 'lofted' into unit cubes. It is easily seen for every cube from layer T1, the portion of T2 that is in contact with this cube is unique. And since there are infinitely many values of α by which we can go on rotating and stacking layers of cubes on top of each other, we can ensure every cube in the full 3D arrangement has a unique neighborhood.
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Remark: Assuming the above argument to be sound, it appears that any shape such as the cube which allows 3-space to be tessellated layer by layer allows such layer-by-layer transformations that ensure every unit has a unique neighborhood. However, I don't know what to do for shapes that do tessellate 3D without allowing layer-by-layer buildup. And of course, in 2D, I have no idea yet as to tiling with a single shape and ensuring unique neighborhoods.

Note: None of the above arrangements - with 2 species of tiles in 2D or with 1 species in 3D - is aperiodic; indeed, they are built of infinite strips (or layers) that are periodic arrangements of identical rectangles (or boxes).