= Solution
Rows in the following matrix index the source, columns the target, in the order $X,Y,Z$. Solving the arrow-square equation for a <quiver representation morphism> gives
$$
\boxed{\bigl(\dim_k\operatorname{Hom}_Q(U,V)\bigr)_{U,V=X,Y,Z}=\begin{pmatrix}1&0&1\\0&1&0\\0&1&1\end{pmatrix}.}
$$
In particular, the nonzero morphisms between distinct representations are \b[$X\to Z$ and $Z\to Y$], each a one-dimensional family of scalar multiples of the vertexwise inclusion or projection. A map $Y\to Z$ is forced to vanish at the source by the identity arrow of $Z$; similarly a map $Z\to X$ is forced to vanish at the target. Maps between $X$ and $Y$ are zero. Each <endomorphism ring> is $k$.
The <extension complex of quiver representations> for the <one-arrow quiver> is
$$
\operatorname{Hom}(U_1,V_1)\oplus\operatorname{Hom}(U_2,V_2)\longrightarrow\operatorname{Hom}(U_1,V_2),\qquad(h_1,h_2)\longmapsto f_Vh_1-h_2f_U.
$$
Its <cokernel> is $\operatorname{Ext}^1_Q(U,V)$. Substitution of the three representations gives
$$
\boxed{\bigl(\dim_k\operatorname{Ext}^1_Q(U,V)\bigr)_{U,V=X,Y,Z}=\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}.}
$$
Here the first argument is the quotient endpoint of a <short exact sequence>. Thus the only possible nonsplit endpoint pair is \b[subobject $X$, quotient $Y$]. The sequence
$$
0\longrightarrow X\longrightarrow Z\longrightarrow Y\longrightarrow0
$$
is nonsplit, since $Z$ is indecomposable. More explicitly, all extensions with these endpoints have a middle arrow $k\xrightarrow{\lambda}k$: $\lambda=0$ gives the <split short exact sequence>, while each $\lambda\ne0$ gives a middle representation isomorphic to $Z$. With endpoint identifications fixed the extension classes form $k$; up to endpoint automorphisms all nonzero classes give this same nonsplit sequence. There are \b[no other nonsplit sequences with the listed endpoints], including equal endpoints.
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