TITLE: Is isomorphism not always unique?
QUESTION [5 upvotes]: Given two isomorphic groups G and H, is it possible that two or more functions define their isomorphism? Also, is it possible that another group say, L is isomorphic to G but not to H?

REPLY [5 votes]: To the first, yes, to the second, no. I'll address the second question first. An isomorphism can be succinctly characterized as a homomorphism $\phi: G \to H$ which is invertible (it is a nice property of homomorphisms and algebraic structure maps in general that the set theoretic inverse of a bijective homorphism is still a homomorphism). If you accept that the composition of two homomorphisms is still a homomorphism, then we we can invert the composition $\psi\phi: G \to L$ of two isomorphisms $\phi: G \to L$, and $\psi: H \to L$ by just taking the reverse composition $\phi^{-1}\psi^{-1}$. In particular, isomorphism is an equivalence relation. 
To the second, the uniqueness of isomorphisms $G \to H$ is basically classified by the automorphisms $$\text{Aut}(G) = \{\psi:G \to G| \psi \text{ an isomorphism}\}$$
It is worthwhile to check that the above set is a group under composition of functions. In particular, given a particular isomorphism $\phi: G \to H$, we can define a new isomorphism $G \to H$ by simply composing with an element of the automorphism group $\phi \circ \psi$. In fact, this is a  bijection, though it is not in any way canonical. There are exactly as many isomorphism $G \to H$ as there are automorphisms of $H$. (This makes some sense if we think of $G,H$ as being essentially the same group up to a relabelling)
A final note, which is the death knell of the uniqueness of isomorphisms except in two very small cases: There are no groups with non-trivial automorphism group except the trivial group $\{e\}$ and $\mathbb{Z}/2\mathbb{Z}$.