Let f be an isomorphism between two ordinals S and T. In other words, f is 1-1 and onto, and respects set membership. We know that one of the ordinals is a subset of the other. Assume T⊆S. If S ≠ T, f squashes S down onto T. Let z be the least element of S such that f(z) ≠ z. Write f(z) = y, where z < y. At the same time, f(x) = z for some x in S, where z < x. Now z < x, yet f(z) > f(x), hence f is not an isomorphism. Isomorphic ordinals are identical.
If S and T are two well ordered sets, and there is an isomorphism between them, it is unique. Let f and g be different isomorphisms from S onto T, and let z be the least element of S such that f(z) ≠ g(z). Specifically, f(z) = x and g(z) = y, where x < y. Now g(w) = x for some w < z. This means f(w) agrees with g(w) = x, hence f maps two elements onto x, and is not an isomorphism after all. The isomorphism between well ordered sets is unique.