$\require{AMScd}$ Let $R$ be a commutative ring and \begin{CD} 0 @>>>A @>f>> B@>g>> C@>>>0 \end{CD} a short exact sequence of left $R$-modules. Let $\phi_A$, $\phi_B$, and $\phi_C$ be module endomorphisms of $A$, $B$, and $C$, respectively, such that the diagram \begin{CD} 0 @>>>A @>f>> B@>g>> C@>>>0\\ _@V \phi_A V V @V \phi_B V V @V \phi_C V V\\ 0 @>>>A @>f>> B@>g>> C@>>>0 \end{CD} commutes. I am trying to prove that $\phi_B =\text{id}_B$ if and only if $\phi_A =\text{id}_A$ and $\phi_C =\text{id}_C$. It is easy to show that if $\phi_B =\text{id}_B$, then $\phi_A =\text{id}_A$ and $\phi_C =\text{id}_C$ using commutativity and the fact that $f$ is injective and $g$ surjective. How might I prove the other direction? I have tried chasing elements but cannot get the result.
The claim may also be false, so a counterexample would be useful as well.
$\require{AMScd}$
The direction you want to prove is false. Good sources of examples are non-split sequences. Having a middle term with non-trivial automorphisms helps. Everything in my counterexample will be in the category of abelian groups.
Consider the diagram \begin{CD} 0 @>>>\mathbb{Z}_2 @>f>> \mathbb{Z}_4@>g>> \mathbb{Z}_2@>>>0\\ _@V 1_A V V @V \phi V V @V 1_C V V\\ 0 @>>>\mathbb{Z}_2 @>f>> \mathbb{Z}_4@>g>> \mathbb{Z}_2@>>>0 \end{CD} where $f(1) =2$, $g(1)=1$, and $\phi(1)=3$. (I need only describe these maps on $1$ since they're all cyclic groups.)
You can check that my sequence is exact: it just realizes the quotient group $\mathbb{Z}_4/\langle 2 \rangle$ as isomorphic to $\mathbb{Z}_2$. You may also check that the entire diagram commutes: $$ \phi (f(1))= \phi(2) = 3+3 = 2= f(1) $$ and $$ g(\phi(1))=g(3)= 3 =1 = g(1). $$ Certainly, $\phi \neq 1_{\mathbb{Z}_4}$.