Theorem:
Let $A \in \mathbb{R}^{n \times n}$ whose (each) column sums to one. Then, $A \vec{X} = \vec{X}$ has non-trivial solution.
Proof:
By definition, A is a left stochastic matrix.
Observe:
$A\vec{X}=\vec{X}$ is a transformation under an identify map.
$\Rightarrow \vec{X} = A^{-1}\vec{X}$
But recall that for any matrix $A \in \mathbb{n \times n}$: $A$ is invertible so the inverse of $A, A^{-1}$ exists.
By the uniqueness theorem, any solution $\vec{X}$ to $A\vec{X}=\vec{b}$ is unique. Clearly, for $\vec{b} = \vec{0}$, $A\vec{X}=\vec{0}$ can be satisfied by $\vec{X} = \vec{0}$ and only $\vec{X} = \vec{0}$.
Since, $A$ is not the zero matrix and $\vec{X} \neq \vec{0}$, $\vec{X} = A^{-1}\vec{X}$ is not the zero vector.
Indeed, no trivial solution exists.
Am I going around in circles in my above attempt?
Any help is greatly appreciated.