My question was addressed here. The answers claims there are two isomorphisms, but I think that is not correct (or I really think I might be wrong, but don't see why).
The homorphisms $\varphi:\mathbb{Z}\rightarrow \mathbb{Z}$ are of the form $\varphi(n)=n\varphi(1)$, that is multiplication by $\varphi(1)$. Since we have the additive group, the identity is $0$. Hence the kernel is given by:
$$\text{ker} \varphi=\{a\in \mathbb{Z}| \varphi(a)=0\}$$
But $\varphi(a)=a\varphi(1)$, then if $a=0\rightarrow \varphi(a)=0 \quad \forall \varphi(1)$, that is all homorphisms are also isomorphisms, which to me makes sense, because ultimately $\varphi$ is a linear map, which is injective and surjective, hence an isomorphism.
I'd like to know if there is a flaw in my reasoning, and this answer is correct, and if so, where is my flaw how to get to that answer using the kernel.
Best Answer
I think bringing in the technical language of kernels has obscured the point. An isomorphism $f:G\rightarrow H$ is a bijective homomorphism. You're right that the homomorphisms $\mathbb{Z}\rightarrow\mathbb{Z}$ are exactly the linear maps; so in order to count the isomorphisms we want to count the bijective linear maps.
A little thought shows that not every linear map is bijective. For example:
$x\mapsto 0\cdot x$ is not injective.
$x\mapsto -2\cdot x$ is not surjective.
So we're left with the question:
And it's not hard to see the answer:
How does this interact with the language of kernels?
Well, calculating the kernel of $f$ will address half of the question of whether $f$ is a bijection: specifically, $f$ is injective iff $ker(f)=\{0\}$. However, it does not tell you anything about surjectivity. So you can't fully understand the isomorphisms just by thinking about kernels.
(Actually that's not quite true: for maps from a finite set to itself, injectivity implies surjectivity - and so a self-homomorphism of a finite group is an automorphism iff its kernel is $\{0\}$. However, that doesn't apply in the $\mathbb{Z}$ case.)