Let $Q$ be a quiver, with dimension vector $d$ and let $e$ be another dimension vector, such that $d_v\leq e_v$ for every vertex $v$ of $Q$. If $M$ is a $K$-representation of $Q$ of dimension vector $d$, it induces a representation $\widetilde{M}$ of dimension vector $e$ in the following way: for every vertex $v$, define the corresponding vector space $\widetilde{M}_v$ to be $\widetilde{M}_v:=M\oplus K^{e_v-d_v}$. The linear maps of $\widetilde{M}$ are given by extending those of $M$ by $0$ on the new summand.
Question: let $M$ and $N$ be $d$-dimensional representations. Assume that $\widetilde{M}\cong \widetilde{N}$. Is it true that $M\cong N$?
I cannot see a reason why this should be true. However, the only examples over which I could test it explicitly are the Dynkin quivers and the loop quiver, and it seems to me that in this cases the answer is indeed positive.
Take the loop quiver for example. Then an isomorphism $\widetilde{M}\cong \widetilde{N}$ does not induce an isomorphism $M\cong N$. However, since similarity may be characterized with "having the same Jordan form over the algebraic closure of $K$", indeed we have $M\cong N$. To me, this seems a subtle question (at least to prove that it is true, but maybe someone sees an immediate counterexample and kills it off immediately...).
