= Solution
We first record why <rigid quiver representations have open orbits>. Let $\mathbf n=\dim X$, $G=\prod_i\operatorname{GL}(X_i)$ and $R=\operatorname{Rep}_Q(\mathbf n)$. The stabilizer is $\operatorname{Aut}_Q(X)$, the nonempty open set of units in $\operatorname{End}_Q(X)$, so it has dimension $\dim\operatorname{End}_Q(X)$. The <dimension formula for an algebraic group homomorphism>, or the same constant-fiber argument for the orbit map, gives
$$
\dim\mathcal O_X=\sum_i n_i^2-\dim\operatorname{End}_Q(X).
$$
By the <Ringel form> identity and $\operatorname{Ext}^1_Q(X,X)=0$, this equals $\sum_\sigma n_{s(\sigma)}n_{t(\sigma)}=\dim R$. An algebraic-group orbit is locally closed; since $R$ is an irreducible <affine space>, this full-dimensional orbit is open and dense.
Write $q=\rho\pi$ for the composed path. Each matrix entry of $q$ is a polynomial function on $R$. Since $qX=0$, change of basis makes it zero on all of $\mathcal O_X$, hence on all of $R$ by density. Suppose instead that $X_{t(\rho)}\ne0$. The condition $\pi X\ne0$ ensures that every vertex space visited by $\pi$ is nonzero. Choose a vector $v_a\ne0$ and a functional $\ell_a$ with $\ell_a(v_a)=1$ at every vertex visited by $q$. Assign to every arrow $\sigma$ occurring in $q$ the map $v_{t(\sigma)}\ell_{s(\sigma)}$, and assign arbitrary maps, say zero, to the other arrows. At this representation, the path $q$ carries its initial chosen vector to its final chosen vector, so $q$ is nonzero, a contradiction.
The construction uses a single assigned map per arrow, so it still works if an arrow or vertex occurs repeatedly in the path. Therefore
$$
\boxed{X_{t(\rho)}=0.}
$$
This proves the <path identities in a rigid quiver representation> claim for an arbitrary <quiver>. Maximal rank of the individual arrows alone would not justify the conclusion about their composition.
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