Are exact sequence homomorphisms unique? Oftentimes I see examples short exact sequences written as $0 \to A \to B \to C \to 0$ without explicitly saying what the morphisms are. Of course the definition requires ${\rm im}(g) = \ker(f)$, but a lot of times in text the functions are omitted. For example, $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$ is a famous example of a non-split exact sequence, yet the morphisms are implicit or an exercise for the reader.
This leads me to wonder whether, fixing $A$, $B$, and $C$, are the morphisms uniquely determined? Or do there exist groups (or another algebraic structure) $A$, $B$, $C$ such that there are multiple distinct short exact sequences of the form $0 \to A \to B \to C$ with differing morphisms?
 A: It's entirely possible to have multiple distinct short exact sequences. For a cheap example, you can take any (nontrivial) automorphism $\varphi$ of $B$. Then if we have a short exact sequence
$$
0 \to A \overset{\iota}{\to} B \overset{\pi}{\to} C \to 0
$$
we get a new short exact sequence by composing with $\varphi$:
$$
0 \to A \overset{\varphi \circ \iota}{\to} B \overset{\pi \circ \varphi^{-1}}{\to} C \to 0
$$
You might reasonably complain about this, and ask if there are any less silly examples. Here the answer is also yes, but we need to make explicit what we mean by "less silly".
Say that two extensions
$0 \to A \overset{\iota}{\to} B \overset{\pi}{\to} C \to 0$
and
$0 \to A \overset{\iota'}{\to} B' \overset{\pi'}{\to} C \to 0$ are equivalent extensions iff there is an isomorphism $\varphi : B \cong B'$ so that the following diagram commutes.

Then it's possible for $B$ and $B'$ to be isomorphic as mere abelian groups, but not equivalent as extensions!
The quickest way to see this requires a very cool fact that you may not know: The set of extensions of $C$ by $A$ up to equivalence forms a group, called $\text{Ext}^1(C,A)$. It turns out to be routine (with practice) to calculate these groups in simple cases -- this is the subject of homological algebra.
Then we can compute, for instance, $\text{Ext}^1(\mathbb{Z}/p, \mathbb{Z}/p) \cong \mathbb{Z}/p$. This tells us that there are $p = |\mathbb{Z}/p|$ many inequivalent extensions $E$ fitting into a diagram
$0 \to \mathbb{Z}/p \to E \to \mathbb{Z}/p \to 0$.
Of course, there are only two possible choices of $E$ up to isomorphism, since it must be an abelian group of order $p^2$! Either $E \cong (\mathbb{Z}/p) \oplus (\mathbb{Z}/P)$ or $E \cong \mathbb{Z}/(p^2)$. In particular, when $p \neq 2$, there are more classes of inequivalent extensions than there are possible middle terms! So we must have $E$ and $E'$ which are isomorphic abstractly but not equivalent as extensions!
In a situation like this, writing $0 \to \mathbb{Z}/p \to \mathbb{Z}/(p^2) \to \mathbb{Z} \to 0$ is ambiguous, because we don't know which extension is meant! For instance, if $p \neq 2$ then
$$
0 \to \mathbb{Z}/p \overset{1 \mapsto p}{\longrightarrow} \mathbb{Z}/(p^2) \to \mathbb{Z}/p \to 0
$$
$$
0 \to \mathbb{Z}/p \overset{1 \mapsto 2p}{\longrightarrow} \mathbb{Z}/(p^2) \to \mathbb{Z}/p \to 0
$$
are inequivalent extensions! Of course, in practice, almost nobody writes this down explicitly, since the extension in question is usually "clear from context" once you get used to things.

I hope this helps ^_^
