Endomorphism ring of indecomposable representations Let $Q$ be the quiver given by an $n\times n$ grid where every square commutes and let $F:Q\to {\rm vec_k}$ be an indecomposable (finitely dimensional) representation of $Q$. 
I am interested in examples of such $F's$ with endomorphism ring larger than $k$.
(In the grid all horizontal arrows point in the same direction, and likewise for vertical ones.)
 A: I'll describe one way to construct, starting with any representation of $k[t]$, a representation of $Q$ (for $n=4$) which has the same endomorphism algebra. Since the two-dimensional representation of $k[t]$ where $t$ acts by $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is indecomposable with two-dimensional endomorphism algebra, this gives an example of what you want.
Let $V$ be a representation of $k[t]$, where $t$ acts by $\alpha$.
First, construct a representation of the quiver
$$\require{AMScd}\begin{CD}
@.\bullet @.\\
@. @VVV @. \\
\bullet @>>> \bullet @<<< \bullet\\
@. @AAA @.\\
@. \bullet @.
\end{CD}$$
as follows
$$\begin{CD}
@.V @.\\
@. @VV\beta_1V @. \\
V @>\beta_4>> V\oplus V @<\beta_2<< V\\
@. @AA\beta_3A @.\\
@. V @.
\end{CD},$$
where $\beta_1=\begin{pmatrix}1\\0\end{pmatrix}$, $\beta_2=\begin{pmatrix}0\\1\end{pmatrix}$, $\beta_3=\begin{pmatrix}1\\1\end{pmatrix}$, $\beta_4=\begin{pmatrix}1\\\alpha\end{pmatrix}$.
Then it's easy to check that the endomorphism algebra is the same as that of $V$ as a $k[t]$-module (note that $\beta_1$ and $\beta_2$ fix two two-dimensional subspaces of $V\oplus V$, $\beta_3$ fixes an isomorphism between them that allows us to identify them, and $\beta_4$ then gives an endomorphism of the identified space). 
Finally, given any representation
$$\begin{CD}
@.U_1 @.\\
@. @VV\gamma_1V @. \\
U_4 @>\gamma_4>> W @<\gamma_2<< U_2\\
@. @AA\gamma_3A @.\\
@. U_3 @.
\end{CD}$$
of that quiver, the representation
$$\begin{CD}
0@>>>0@>>>0@>>>U_1\\
@VVV @VVV @VVV @VV\gamma_1V\\
0@>>>0@>>>U_2@>\gamma_2>>W\\
@VVV@VVV@VV\gamma_2V@|\\
0@>>>U_3@>\gamma_3>>W@=W\\
@VVV@VV\gamma_3V@|@|\\
U_4@>\gamma_4>>W@=W@=W
\end{CD}$$ 
of $Q$ has the same endomorphism algebra.
