Let $Q$ be a linear Dynkin quiver of type $A_n$, which looks like
$$1 \longrightarrow 2 \longrightarrow 3 \longrightarrow \cdots \longrightarrow (n-1) \longrightarrow n$$
Then we know by Gabriel's theorem that the indecomposable representations of $Q$ are given by $M^{r,s}=(M^{r,s}_i,M^{r,s}_a)_{i \in Q_0, a \in Q_1}$ where $M^{r,s}_i=K$ for $r \leq i \leq s$ and $M^{r,s}_i=0$ otherwise.
Now $Q$ has two subquivers of type $A_{n-1}$ with indecomposable representation denoted by $N^{r,s}=(N^{r,s}_i,N^{r,s}_a)$ for $1 \leq r \leq s \leq n-1$. Is it true that for one subquiver it holds $N^{r,s} \cong M^{r,s}$ for all $1 \leq r \leq s \leq n-1$ and for the other $N^{r,s} \cong M^{r,s}$ for all $2 \leq r \leq s \leq n$? Intuitively I would say yes, it makes sense. But I still have some doubts on whether it is possible to make this statement.
So yeah, it's not quite right to say they're isomorphic since $M^{r,s}$ and $N^{r,s}$ are representations of different quivers. You're morally right though. To make it more precise you've gotta say something like, looking at the "left" case, you've got a morphism of quivers $\phi\colon A_{n-1} \to A_n$ that maps the nodes of $A_{n-1}$ to the nodes labelled $1, \dotsc, n-1$ of $A_n$. This morphism of quivers $\phi$ induces a functor on the representation categories $\Phi \colon \mathrm{Rep}A_n \to \mathrm{Rep}A_{n-1}$ where $\Phi(M^{r,s})_i = M^{r,s}_{\phi(i)}$ and $\Phi(M^{r,s})_a = M^{r,s}_{\phi(a)}$. Essentially you're just forgetting about that last node and arrow of $A_n$ to get a representation of $A_{n-1}$. The big idea is that since $\phi$ is injective, then $\Phi$ is surjective, and $N^{r,s}$ is isomorphic to $\Phi(M^{r,s})$ for every $r$ and $s$ such that $1 \leq r\leq s < n$.