I was reading 'Introductory real analysis' by Kolmogorov and Fomin and it mentioned in section 3.3 that "ordered sets with same power need not necessarily have same type of order" A counter example given is that set of positive integers (which obviously have same power with itself) can be ordered differently by a) natural way of ordering, a<=b iff a-b is non positive and b)1<3<5....<2<4<6... i.e with additional rule that all odd numbers are 'smaller' than even numbers
it seems trivial at first that a mapping where every element to itself is not an isomorphism,since 2<3 in first order but 2>3 in the second. Then a second thought comes in that: can there be any other type of mapping that is isomorphism? Just showing that identical mapping is not an isomorphism is not enough to show that those two orders are 'different', I need to show that no possible isomorphisms can exist. In general, how do I prove that there can not be an isomorphism here?
Yes, you are correct, showing that just one map is not an isomorphism is not enough.
But you can show that in the natural ordering every element has finitely many predecessors; and that this property of having finitely many predecessor is something preserved by isomorphisms. In the new ordering define, $2$ has infinitely many predecessors.
Generally, showing that two structures (linear or partial orders, groups, fields, etc.) are not isomorphic can be, and perhaps should be, attacked by trying to find some property which is preserved under isomorphism and holds only in one structure. For example having a minimal element for orders; being commutative for groups; having $x$ such that $x^2=-1$ for fields. All those are properties which can be used to "separate" the isomorphism types of structures.