(Short) Exact sequences with no commutative diagram between them

$\def\ZZ{\mathbb{Z}}$ This can happen in finitely generated abelian groups. Let $p$ be prime, set $G = \ZZ/p^2 \oplus \ZZ/p$ and set $H = \ZZ/p^3 \oplus \ZZ/p^2 \oplus \ZZ/p$. Then there are two non-isomorphic short exact sequences $0 \to G \to H \to G \to 0$. The first one is the sum of the extensions: $$\begin{matrix} 0 & \to & \ZZ/p & \to & \ZZ/p^3 & \to & \ZZ/p^2 &\to& 0 \\ 0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 & \to & 0 &\to& 0 \\ 0 & \to & 0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 \\ \end{matrix}$$ and the second is the sum of $$\begin{matrix} 0 & \to & \ZZ/p^2 & \to & \ZZ/p^3 & \to & \ZZ/p &\to& 0 \\ 0 & \to & 0 & \to & \ZZ/p^2 & \to & \ZZ/p^2 &\to& 0 \\ 0 & \to & \ZZ/p & \to & \ZZ/p & \to & 0 & \to & 0 \\ \end{matrix}$$

Let's see that these are not isomorphic. Write $\alpha$ for the map $G \to H$. In the first extension, $\alpha(p G) \cap p^2 H = (0)$; in the second extension, $\alpha(pG) = p^2 H$.

In any reasonable category, isomorphism classes of extensions $0 \to X \to ?? \to Z \to 0$ are classified by the orbits of $\mathrm{Aut}(X) \times \mathrm{Aut}(Z)$ on $\mathrm{Ext}^1(Z,X)$. You are asking for cases where there is more than one orbit which makes the center term isomorphic as an abstract element of the category. There is no reason this shouldn't happen, so I would expect it to happen basically any time there are nontrivial extension groups available.

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To add some more high level context, the analogous example works in $k[t]$-modules. (That is to say, extending $k[t]/t^2 \oplus k$ by itself to get $k[t]/t^3 \oplus k[t]/t^2 \oplus k$.) In general, for any partition $\lambda$, let $M(\lambda)$ be the $k[t]$ module $\bigoplus_i k[t]/t^{\lambda_i}$. Then $\mathrm{Ext}^1(M(\mu), M(\lambda))$ is stratified into locally closed pieces according to the isomorphism type of the extension, and the number of components where the extension is isomorphic to $M(\nu)$ is the Littlewood-Richardson number $c_{\lambda \mu}^{\nu}$. This example is $c_{(21) (21)}^{(321)}=2$. Since $Aut(M(\lambda))$ is always a connected group (it is a unipotent extension of $\prod GL(\lambda_i)$), it can't mix the components.

You can start to get context for this from the first two chapters of Schiffmann's Lectures on Hall Alegbras. Unfortunately, Schiffmann always takes $k$ to be a finite field so that he can count points, which makes it harder to talk about positive dimensional subvarieties of $\mathrm{Ext}^1$.

Others have given examples. It might be worth noting that (in the category of finitely generated abelian groups, or more generally in the category of finitely generated modules over a noetherian ring) there can be no example with $X_2$ isomorphic to $X_1\oplus X_3$ .

Indeed suppose we have such a sequence $$0\rightarrow X_1\rightarrow X_2\rightarrow X_3\rightarrow 0$$

We want to show that any such sequence splits, so that all such sequences are isomorphic.

Apply $Hom(,X_1)$ to get

$$0\rightarrow Hom(X_3,X_1)\rightarrow Hom(X_2,X_1)\rightarrow Hom(X_1,X_1)$$

The original sequence splits if and only if this sequence is surjective on the right. For this, it suffices to check surjectivity after localizing and then completing at an arbitrary prime $P$ and for this it suffices to check exactness after modding out an arbitrary prime power $P^n$. Now everything is of finite length so to get surjectivity, all we need is for the lengths on the left and right to add up to the length in the middle, which is automatic when $X_2\approx X_1\oplus X_3$.

It is possible to reuse my answer to this question. It gives an example in the category of $\mathbb C[x,y]$-modules.

Let $R$ be the ring $R=\mathbb C[x,y]$, and let $B$ be the $5$-dimensional $R$-module with shape like a 'W'. That is, basis elements are $a_1,a_2,a_3,b_1,b_2$ and the module structure is given by $$y \cdot a_1=b_1,$$ $$x \cdot a_2=b_1,$$ $$y \cdot a_2=b_2,$$ $$x \cdot a_3=b_2,$$ and all other products of generators and basis elements are zero.

Let $A=\mathbb C$ be the trivial $R$-module and consider the parallel morphisms $f,f' \colon A \rightarrow B$ defined by $f(z)=zb_1$ and $f'(z)=zb_2.$ Now ${\mathrm{coker}} \; f \simeq {\mathrm{coker}} \; f'$ as $R$-modules, but $f$ and $f'$ are non-isomorphic in $\mathrm{Mor}(\mathrm{Mod} \; R)$.

That $f$ and $f'$ are non-isomorphic in $\mathrm{Mor}(\mathrm{Mod} \; R)$ is another way of saying there is no commutative diagram $$ \begin{array}{ccc} A & \xrightarrow{f} & B \\ \downarrow u & & \downarrow v \\ A & \xrightarrow{f'} & B \end{array} $$ with $u$ and $v$ isomorphisms. Here $f$ and $f'$ are monomorphisms, so we obtain an example satisfying the conditions given in the question.