This is an exercise from An Introduction to Quiver Representations by Harm Derksen and Jerzy Weyman.
Exercise 1.6.3. Let $Q$ be the quiver \begin{equation*} \circ \longrightarrow \circ \longleftarrow \circ \end{equation*} and let $A=kQ$. Show that $A$ is not isomorphic to $A^{\text{op}}$.
Recall that $A^{\text{op}} = kQ^{\text{op}}$, where $Q^{\text{op}}$ is the quiver obtained by reversing the arrows. My first attempt was to look for some algebraic invariant related to a maximal complete set of orthogonal idempotents, but I can't see it clear.
In order from left to right, name the three idempotents $e_1$, $e_2$, and $e_3$. For any $p \in \mathbf{k}Q^\text{op}$ only $e_2$ (and multiples of it) has the property that $e_2 p = k e_2$ for some $k \in \mathbf{k}$. But quite a few elements of $\mathbf{k}Q$ have this property, like $e_1$ and $e_3$ and any linear combination of them. You can use this to arrive at a contradiction to there being an isomorphism $\mathbf{k}Q \cong \mathbf{k}Q^\text{op}$.