Categoricity: the natural numbers are unique up to unique isomorphism

Apply the recursion theorem [L1] in $(N,0,\sigma )$ with $A=N^{\mathrm{\prime }}$, $a=0^{\mathrm{\prime }}$, $f={\sigma }^{\mathrm{\prime }}$: there is a unique $h:N\to N^{\mathrm{\prime }}$ with $h(0)=0^{\mathrm{\prime }}$ and $h(\sigma (n))={\sigma }^{\mathrm{\prime }}(h(n))$.

Symmetrically, apply [L1] in $(N^{\mathrm{\prime }},0^{\mathrm{\prime }},{\sigma }^{\mathrm{\prime }})$ with $A=N$, $a=0$, $f=\sigma$: there is a unique $h^{\mathrm{\prime }}:N^{\mathrm{\prime }}\to N$ with $h^{\mathrm{\prime }}(0^{\mathrm{\prime }})=0$ and $h^{\mathrm{\prime }}({\sigma }^{\mathrm{\prime }}(n^{\mathrm{\prime }}))=\sigma (h^{\mathrm{\prime }}(n^{\mathrm{\prime }}))$.

The composite $h^{\mathrm{\prime }}\circ h:N\to N$ satisfies $(h^{\mathrm{\prime }}\circ h)(0)=h^{\mathrm{\prime }}(0^{\mathrm{\prime }})=0$ and $(h^{\mathrm{\prime }}\circ h)(\sigma (n))=h^{\mathrm{\prime }}({\sigma }^{\mathrm{\prime }}(h(n)))=\sigma (h^{\mathrm{\prime }}(h(n)))=\sigma ((h^{\mathrm{\prime }}\circ h)(n))$; the identity ${\mathrm{i}\mathrm{d}}_N$ satisfies the same two equations, so by induction [L2] the set $\{n:(h^{\mathrm{\prime }}\circ h)(n)=n\}$ is all of $N$, i.e. $h^{\mathrm{\prime }}\circ h={\mathrm{i}\mathrm{d}}_N$.

Likewise $h\circ h^{\mathrm{\prime }}={\mathrm{i}\mathrm{d}}_{N^{\mathrm{\prime }}}$ by the same argument carried out in $N^{\mathrm{\prime }}$.

Hence $h$ is a bijection with inverse $h^{\mathrm{\prime }}$, and it preserves the base point and the successor, so it is an isomorphism of Peano systems.

Uniqueness: any $h_1:N\to N^{\mathrm{\prime }}$ with $h_1(0)=0^{\mathrm{\prime }}$ and $h_1(\sigma (n))={\sigma }^{\mathrm{\prime }}(h_1(n))$ satisfies the recursion clauses defining $h$ in step 1.1, so $h_1=h$ by the uniqueness in [L1] (equivalently $\{n:h_1(n)=h(n)\}$ contains $0$ and is closed under $\sigma$, hence is all of $N$ by induction [L2]).