Let $G$ be a graph with $n$ vertices and let $\chi_G$ be its chromatic polynomial.
Since $\chi_G(q) = q(q-1)^{n-1}$ for trees, we have that the linear coefficient is of unit modulus.
Is this condition also sufficient for a graph to be a tree? If so, how to prove this?
Thank you.
It is sufficient. It is known that the absolute value of the linear coefficient of the chromatic polynomial counts the number of acyclic orientations of $G$ with some fixed $v \in V$ as their unique source, see the answer to this question.
Your statement easily follows from this: first, a non-connected graph has no such orientations, since each component must contain at least one source. Second, a connected graph that is not a tree has multiple such orientations: take a spanning tree of $G$ that contains every neighbour of $v$ and orient the edges of this tree away from $v$. Then we can orient the rest of the edges arbitrarily to get a suitable orientation, and since we can orient each extra edge in two ways, this method already gives us multiple orientations.