Isomorphisms between cyclic groups!

Ok, here is something i thought i understood, but it turns out i am having difficulties fully grasping/proving it.

Let $$\theta:G->G'$$ be an isomorphism between G and G', where o(G)=m=o(G'), and both G and G' are cyclic, i.e. G=[a] and G'=

So my question is, when we want to find the total number of isomorphisms from G to G', i 'know' that the total number of such isomorphisms is basically
the number of generators of G':
$$\theta(a)=b^k$$ where gcd(k,m)=1. But i don't really understand why? That is, how are we sure that by just counting the number of generators on G' we are actually finding the total number of such isomorphisms. ?

Any explanations would be appreciated.

HallsofIvy
Suppose $\theta$ were any isomorphism from G to G'. Apply $\theta$ to a generator of G. It is simple to prove that $\theta$ MUST map a generator to a generator. So it is easy to see that it is a matter of how many generators each has: each mapping of generator to generator gives an isomorphism.
first $$\theta(a)=b^i; \theta(a^3)=b^i, \theta(a^5)=b^i;\theta(a^7)=b^i,i=1,3,5,7$$ , which means not 4, but 4! isomorphisms.