TITLE: How many isomorphisms between two abelian groups
QUESTION [5 upvotes]: I'm trying to work out how many isomorphisms there are from $V_4$ (Klein-4) to $C_2 \times C_2$ (2nd cyclic group)
Firstly, I've already shown they are indeed isomorphic and clearly there are $4!$ bijections from $V_4$ to $C_2 \times C_2$, however only $4!/4 = 6$ of these map the identity to the identity, so there are at most $6$ isomorphisms from $V_4$ to $C_2 \times C_2$.
Now, I'm slightly confused at this point, mostly because both are abelian. The map I used to show they are isomorphic I 'constucted' by comparing the Cayley tables of both, and showing they are 'in essence' exactly the same, and I mapped element to element in the order they appeared on the Cayley table. My confusion comes in because I'm not sure of the consequences of mapping the non-identity elements differently, so ultimately don't really know whether there are $6$ isomorphisms or just $1$. Having said that, I did map the non-identity elements differently and saw that it's still an isomorphism, that is $\phi (g_1 g_2) = \phi (g_1) \phi(g_2)$ for $g_1, g_2 \in V_4$. So I'm thinking that there are $6$ isomorphisms, but because I'm doubtful I thought I'd post it for some insight.
Thanks a lot.

REPLY [2 votes]: Consider the isomorphisms going one way, all arrows to the same object. The comparison of those objects is an automorphism. The question is how you tell that those are two isomorphisms to the same object. So, the distinctness of isomorphisms is measured by needing a non-identity automorphism to convert the two. So, there can be no more isomorphisms than there are automorphisms of the target. However, the composition of isomorphisms is an isomorphism, so composing any isomorphism with an automorphism of the target will yield a new isomorphism to the target. Hence, the number of isomorphisms between two objects is the same as the number of automorphisms of either object.