My solution:
$C4$ = {$e,r,r^2,r^3$} where e-identity element and r is rotation by 90°
$Z4$ = {$0,1,2,3$}
$|Z4|=|C4|=4$
We see that both groups are cyclic.
$C4$ is generated by $r$ and $r^3$.
$Z4$ is generated by $1$ and $3$.
We know that cyclic groups of same order are isomorphic.
We know that isomorphism preserves the order of the element.
Order of the elements:
$|e|=1 ------- |0|=1$
$|r|=4 -------|1|=4$
$|r^2|=2 -------|2|=2$
$|r^3|=4 ---- ---|3|= 4$
We see that in both groups there is 1 element with order equal 1, 1 element with order equal 2 and 2 elements with order equal 4.
So we can create 2 mappings:
ψ : C4 -> Z4
$ψ(e)=0$
$ψ(r)=1$
$ψ(r^2)=2$
$ψ(r^3)=3$
ϕ: C4 -> Z4
$ϕ(e)=0$
$ϕ(r)=3$
$ϕ(r^2)=2$
$ϕ(r^3)=1$
Both mappings are of course bijective because they are 1-1 and onto.
Now we have to prove that ψ and ϕ are homomorphisms.
Let's begin with $ϕ$ :
Let $\phi(r^m)=3m$
$\phi(r^mr^n) = \phi(r^{m+n}) = 3(m+n) = 3m + 3n = \phi(r^m) + \phi(r^n)$.
So ϕ is homomorphism and a bijection so ϕ is isomorphic.
Now we do the same with ψ:
Let $\psi(r^a)=a$
$\psi(r^ar^b) = \psi(r^{a+b}) = a + b =\psi(r^a) + \psi(r^b)$.
So $ψ$ is homomorphism and a bijection so $ψ$ is isomorphic.
We have proved that there exist 2 isomorphisms between $Z4$ and $C4$.