A question concerning binary operations and isomorphisms Consider two nonempty sets $A$ and $B$, such that $A$ is isomorphic to $B$. Now, if $B$ is a group under some binary operation $*$, does it necessarily imply that there exists an operation $*'$, under which $A$ is a group? If this is true, what happens if I strengthen my condition, saying for instance that $B$ is an abelian group under $*$? Would there necessarily be a $*'$ under which $A$ is abelian?
The reason I ask this is because of an interesting question I came across which I've been trying to answer for days (please do not answer it, if you happen to know the answer); 

Consider $S = \mathbb N \cup \{0\}$. Is it possible to make $S$ an
  abelian group under some binary operation?


If there is a bijection $f\colon A\to B$ and $*$ is a group operation on $B$, you can define, for $x,y\in A$,
$$
x*'y=f^{-1}(f(x)*f(y))
$$
This defines an operation on $A$ and $f$ becomes an isomorphism between $(A,*')$ and $(B,*)$, since, by definition,
$$
f(x*'y)=f(x)*f(y)
$$
and $f$ is bijective. It's just routine to check that $(A,*')$ is a group if $(B,*)$ is a group; abeliannes is preserved.
