Let . It suffices to show that the three conditions work for , for a translation trivially doesn't change any of the properties we wish to prove about .

For each , consider the parallelogram formed by , , , and . Remark that the distance between any of these two points is preserved under , so it's not hard to show that the shape formed by , , , and is also a parallelogram. Furthermore, remark that by the definition of , we have . Therefore by parallelogram addition .

Now consider the points , , and , where and . Remark that these three points are collinear, and so by the equality case of the triangle inequality , , and are also collinear. Furthermore, implies . Combining these yields .

By the above few paragraphs, therefore, we have that is a linear transformation. Let be the matrix corresponding to . It's well known that if is length-preserving, then is orthogonal. Since any orthogonal matrix is invertible, has full rank, meaning that is bijective, proving part (1).

I believe the fact that is a linear transformation is enough to prove (2) and (3), but I'm not sure, so I'll prove something stronger about . I will actually attempt to characterize all possible functions that can take. Let . Then , so computation gives This means we need and . Let , , , and for some and between and . Then If , then and , while in the other case and . Therefore is of the form which implies that consists of a composition of a rotation and a reflection across the -axis. From here parts b) and c) are trivial.