Why N is their only model (up to isomorphism) in peano system?

Definiton(Natural number set):

The set $N$ is nonempty. $0 \in N$, $S:N \to N$, if they meet the following condition:

1. $S$ is injection , i.e. $S(y_1) = S(y_2) \Rightarrow y_1=y_2$
2. $0 \notin S(N)$
3. $N = \{0\} \cup S(N)$
4. $M$ is a subset of $N$, if $0\in M$, and $\forall_{x \in M}(S(x) \in M)$ then $M = {N}$.

then we call $(N, S, 0)$ a natural number set.

My teacher or some books about set theory said that any two natural number sets satisfy the following theorem, so they are equivalent or isomorphic.

Theorem$(N_1, S_1, 0_1)$ and $(N_2, S_2, 0_2)$ is two natural number sets, then there is a unique bijection $f: N_1 \to N_2$ satisfies

1. $f(0_1) = 0_2$
2. $\forall_{x \in N_1} f(S_1(x)) = S_2(f(x))$

Commutative diagrams:$\require{AMScd}$\begin{CD}N_1 @>S_1>> N_1\\@V f V V @VV f V\\N_2 @>>S_2> N_2\end{CD}

$\require{AMScd}$\begin{CD}x @>S_1>> S_1(x)\\@V f V V @VV f V\\f(x) @>>S_2> f(S_1(x))=S_2(f(x))\end{CD}

I can understand why this theorem is correct.

But I can't understand why this theorem implies that the two natural number sets are equivalent or isomorphic.Why is the definition of two natural number sets isomorphic or equivalent like this?

I know there's the concept of isomorphism in group theory, but natural numbers are not groups. Does the isomorphism here have relationship with the isomorphism in group theory?