Let $Q$ be a quiver $1 \to 2 \to 3$. How to determine whether the following two representations are isomorphic, in other words, how to find invertible matrices $Q_1$, $Q_2$, $Q_3$ such that the following diagram commutes: \begin{matrix} & K^n & \overset{A}{\longrightarrow} & K^m &\overset{B}{\longrightarrow} & K^r \\ & \downarrow_{Q_3} & & \downarrow_{Q_2} & & \downarrow_{Q_1} \\ & K^n & \overset{I_A}{\longrightarrow} & K^m & \overset{I_B}{\longrightarrow} & K^r \end{matrix} where $I_A$ is a diagonal matrix whose upper left $r(A)\times r(A)$ ($r(A)$ denotes the rank of a matrix $A$) block is the identity matrix and all other entries are zero, $n,m,r\in \mathbb{Z}_{\geq 1}$.
For an example: \begin{matrix} & K^3 & \overset{A}{\longrightarrow} & K^2 &\overset{B}{\longrightarrow} & K \\ & \downarrow_{Q_3} & & \downarrow_{Q_2} & & \downarrow_{Q_1} \\ & K^3 & \overset{I_A}{\longrightarrow} & K^2 & \overset{I_B}{\longrightarrow} & K \end{matrix} Find $Q_1 \in K\backslash\{0\}$, an $2\times 2$ invertible matrix $Q_2$, an $3\times 3$ invertible matrix $Q_3$, such that \begin{align*} Q_1 B =I_B Q_2, \quad Q_2 A = I_A Q_3. \end{align*}
My idea is as follows: there exist $Q_1 \in K\backslash\{0\}$ and an $2\times 2$ invertible matrix $Q_2$ such that $Q_1BQ^{-1}_2$ is equivalent to $I_B$. So \begin{matrix} & K^3 & \overset{A}{\longrightarrow} & K^2 &\overset{B}{\longrightarrow} & K \\ & \downarrow_{Q_3} & & \downarrow_{Q_2} & & \downarrow_{Q_1} \\ & K^3 & \overset{Q_2 A Q^{-1}_3}{\longrightarrow} & K^2 & \overset{I_B}{\longrightarrow} & K \end{matrix}
I am stuck whether there exists an invertible matrix $Q_3$ such that \begin{align*} Q_2 A Q^{-1}_3 = I_A. \end{align*}
Who can give me a hint？ Thanks in advance.
I would make use of the classification of indecomposable representations of this quiver. As you may well know, it has six indecomposable representations up to isomorphism, with dimension vectors 100, 010, 001, 110, 011, and 111. (I can add more details if this is unfamiliar.)
The top representation splits up as min($r(A),r(B)$) copies of the representation with dimension vector 111, max($r(A),r(B)$)-min($r(A),r(B)$) copies of either 110 or 011 (depending on which of $r(A), r(B)$ is larger), and enough copies of each simple representation to get to the right dimension.
Then determine the number of copies of each of the indecomposable representations in the bottom representation.
Edited to add: in the example which you added, it isn't necessarily true that the two representations are isomorphic, but suppose they are. The map $I_A$ has a one-dimensional kernel, so $A$ must as well. An isomorphism would have to send the kernel of $A$ to the kernel of $I_A$. That gives you a first step towards defining $Q_3$. Next, I would consider the kernel of $I_B$.
Note that there is no reason to expect that the matrices $Q_i$ should be integer-valued -- they would normally be matrices over $K$.
Edited to add: It is certainly not true that the two representations will be isomorphic for an arbitrary choice of $A$ and $B$.
The following example is wrong (as the OP kindly points out in comments): For an extreme example, let $A$ and $B$ be the zero matrices (of the appropriate sizes). Then that representation would be isomorphic to 3 copies of the simple representation at vertex 1, two copies of the simple at vertex 2, and one copy of the simple representation at vertex 3. Since quiver representations satisfy Krull-Schmidt, so that a module has a unique expression as a sum of indecomposable modules, this rules out the possibility that it is isomorphic to the other representation you gave. (Though it could also be instructive to prove directly that in this case, there are no matrices $Q_i$ which would give an isomorphism.)
Here is one that works: Consider the direct sum of the representations with dimension vectors 110 and 011. The maps $A$ and $B$ are both rank 1, so $$I_A=[1\ 0],\ I_B=\left[ \begin{array}{c}1\\0\end{array} \right]$$
For $110\oplus 011$, the composition of the two maps is zero, while the for the representation defined by $I_A$ and $I_B$, the composition of the two maps is non-zero. The property of the composition being zero or non-zero will be preserved by isomorphisms, so the representations are not isomorphic.