From the documentation page:

[m00x + m01y + m02] = x'
[m10x + m11y + m12] = y'

So if I want to map (x,y) -> (x,y), the identity transform, I want
m00 == m11 = 1
and
m01 == m02 == m10 == m12 = 0.

public affine(m00: Number, m01: Number, m02: Number, m10: Number, m11: Number, m12: Number) : Affine

Therefore, the call should be

affine(1,0,0,0,1,0) But, this causes the shape to disappear!

FX.println(Transform.affine(1,0,0,0,1,0).impl_getAffineTransform());

prints

AffineTransform[[1.0, 0.0, 1.0], [0.0, 0.0, 0.0]]

It seems the java implementation object is getting its arguments passed in the wrong order (column-first instead of row first) when it is created by its javafx peer.

FX.println(Transform.affine(1,2,3,4,5,6).impl_getAffineTransform());

we expect

AffineTransform[[1.0, 2.0, 3.0], [4.0, 5.0, 6.0]]

but instead, get

AffineTransform[[1.0, 3.0, 5.0], [2.0, 4.0, 6.0]].

The correct function documentation corresponding to the actual current implementation is

public affine(m00: Number, m10: Number, m01: Number, m11: Number, m02: Number, m12: Number) : Affine

affine(1,0,0,1,0,0) gives the identity transform, as expected.