Question
Given the sequences of $R-$modules
$$E_1 = 0\to A\to B\to C\to 0$$
$$E_2 = 0\to A'\to B'\to C'\to 0$$
such that $E_1$ is exact and $A\cong A'$, $B\cong B'$ and $C\cong C'$, when is it true that $E_2$ is exact?
Example
Let $A = \mathbb{R}/\mathbb{Z}$. I want to show that $A$ is injective in a particular way. Consider the short exact sequence
$$0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}\to0$$
where $\phi:\mathbb{Z}\to\mathbb{Z}$ is defined by $\phi(k)=nk$ and $\pi:\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z}$ is the natural projection. I know that $A$ is injective if the following short sequence is exact:
$$0\to Hom_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},A)\to Hom_{\mathbb{Z}}(\mathbb{Z},A)\to Hom_{\mathbb{Z}}(\mathbb{Z},A)\to 0.$$
Since $Hom_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},A) = \mathbb{Z}/n\mathbb{Z}$ and $Hom_{\mathbb{Z}}(\mathbb{Z},A) = A$, we have the short exact sequence
$$0\to \mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}\to0$$
which places us in the position outlined above. Therefore, my question is this. Is it valid to say that because this sequence is exact, my target sequence is exact?
This is certainly not true without more hypotheses. For instance, let $E_1$ be any short exact sequence, and let $A'=A$, $B'=B$, and $C'=C$. Let $E_2$ be the sequence obtained by taking the zero maps between these modules. Then $E_2$ is not exact (unless $A$, $B$, and $C$ are all $0$).
It is true if you assume you have isomorphisms that commute with the maps in your sequences. That is, if $E_1$ is $$0\to A\stackrel{f}\to B\stackrel{g}\to C\to 0$$ and $E_2$ is $$0\to A'\stackrel{f'}\to B'\stackrel{g'}\to C'\to 0,$$ you want isomorphisms $h_A:A\to A'$, $h_B:B\to B'$, and $h_C:C\to C'$ such that $f'h_A=h_Bf$ and $g'h_B=h_Cg$. The proof that $E_2$ is exact in this case is a straightforward diagram chase. For instance, if $b\in \ker(g')$, then $g(h_B^{-1}(b))=(h_C^{-1}g'h_B)(h_B^{-1}(b))=h_C^{-1}(g'(b))=0$ so $h_B^{-1}(b)\in\ker(g)$. Since $E_1$ is exact, this means $h_B^{-1}(b)=f(a)$ for some $a\in A$, and then $b=h_B(f(a))=f'(h_A(a))$ is in the image of $f'$.