Consider the Five lemma with abelian groups. If $l$, $m$, $p$, and $q$ are isomorphisms, then $n$ is an isomorphism. Let $n'\colon C\to C'$ be a second homomorphism such that $ n' \circ g=s\circ m$ and $ t \circ n'=p\circ h$. Does $n=n'$ follow?
At first $n'$ has to be an isomorphim, too. If the statement is wrong, I have no counterexample. Often examples with groups like $\mathbb{Z}, \{0\}$ endowed with the addition work, but here I have no intuition. Help would be greatly appreciated. Regards
This boils down to: Must $n$ be the identity in this diagram? $$\require{AMScd} \begin{CD} A @>f>> B @>g>> C @>h>> D @>j>> E \\ @| @| @VVnV @| @| \\ A @>f>> B @>g>> C @>h>> D @>j>> E \\ \end{CD} $$ At least $c-n(c)\in\ker h$ for all $c\in C$, hence $c-n(c)=g(b)$ for some $b\in B$. Since $n\circ g=g$, we conclude $n(c)-n^2(c)=c-n(c)$, i.e., $(n-\operatorname{id})^2=0$. This suggests the following example: $$\def\Z{\mathbb{Z}} \begin{CD} 0 @>>> \Z @>x\mapsto(x,0)>> \Z^2 @>(x,y)\mapsto y>> \Z @>>> 0 \\ @. @| @VVnV @| @. \\ 0 @>>> \Z @>x\mapsto(x,0)>> \Z^2 @>(x,y)\mapsto y>> \Z @>>> 0 \\ \end{CD} $$ with $n(x,y)=(x+y,y)$.