Is Peano's axiom of induction needed to show $n^\prime\ne n^{\prime\prime}$?

This is the statement of Peano's axioms I will assume for this discussion:

1. $$1$$ is a number.
2. To every number $$n$$ there corresponds exactly one number $$n^\prime.$$
3. $$n^\prime=m^\prime\implies n=m.$$
4. $$n^\prime\ne 1$$
5. Let $$P\left[x\right]$$ be a proposition (propositional form) containing the number variable $$x$$. If $$P\left[1\right]$$ holds and if $$P\left[n^\prime\right]$$ follows from $$P\left[n\right]$$ for every number n, then $$P\left[x\right]$$ holds for every number $$x$$.

Is Peano's axiom of induction (axiom #5) needed to show $$n\ne n^\prime$$ for $$n\ne 1$$? That is, nothing in any of the first four axioms individually or in combination seems to preclude the proposition $$n\ne 1\land n= n^\prime.$$

I thought I had shown this by the 5th axiom, but as I typed up this question I realized that my proof needs more work. I'm not asking for the proof. I'm merely asking for a confirmation that it requires the 5th axiom.

The answer seems clearly to be yes. But I've been wrong about such things before.

For the new version of axiom 3, take $$\Bbb N\cup\{*\}$$ with $$x'=x+1$$ for $$x\in\Bbb N$$ and $$*'=*$$.