The Peano axioms of $\Bbb N$ are:

$1.$ $1 \in \Bbb N$, i.e. $\Bbb N$ is not empty and contains an element denoted by $1$.

$2.$ Every natural number has a successor, i.e. $\forall n\in\Bbb N, \exists!s(n)\in\Bbb N$.

$3.$ if $s(n)=s(m)$ then $n=m$.

$4.$ $1\in\Bbb N$ is the only element that is not the successor of a natural number.

$5.$ The axiom of Mathematical Induction is valid:

Let $S\subseteq\Bbb N$ such that

$1)$ $1\in S$

$2)$ $\forall n\in\Bbb N,n\in S\Rightarrow(s(n)\in S)$.

Then $S=\Bbb N$.

I am trying to find an example of a collection "$\Bbb N$'' with 1,2 that satisfies 5 but not 3 and also not 4. (It is easy to find examples satisfying 3 but not 4,5, and 4 but not 3,5. My question is about 5 but not 3,4.) In other words, is there a set "$\Bbb N$'' that has a $1$, successors exist, and induction holds, but $1$ is the successor of an element and also the successor function is not one-to-one? I can't seem to think of an example. I suspect that if 1,2,5 are satisfied, then either 3 or 4 must hold. Is there an elementary proof of this?