$\require{AMScd}$ The "short" Five Lemma concerns the famous form of exact commutative diagram: $$\begin{CD}0@>>>A@>>>B@>>>C@>>>0\\&@VV\simeq V&@VVV&@VV\simeq V&\\0@>>>A'@>>>B'@>>>C'@>>>0\end{CD}.$$ In its weakest form, it says that if the two side maps are isomorphisms are isomorphisms then so is the middle map $B\to B'$ and follows from the normal five lemma by trivially adjoining the $0\to 0$ isomorphisms on each side. However, if I have the following diagram $$\begin{CD}0@>>>A@>>>B@>>>C@>>>0\\&@VVV&@VV\simeq V&@VV\simeq V&\\0@>>>A'@>>>B'@>>>C'@>>>0,\end{CD}$$ is the map $A\to A'$ necessarily an isomorphism, and similarly for $C\to C'$?
I apologize for any "stupid" overlooking on my part - this question feels like it should be completely trivial.
This follows from the "long" 5 lemma, by extending the diagram to have another column of $0$s on the left. Alternatively, it is easy to see directly: if you identify $B$ with $B'$ and $C$ with $C'$ via the isomorphisms, commutativity of the right square says that $B\to C$ and $B'\to C'$ become the same map, and so you get that $A$ and $A'$ are kernels of the same map. Similar arguments apply if you instead have isomorphisms $A\to A'$ and $B\to B'$.