Take an $n \times n$ matrix $A$, and suppose that $v$ is an eigenvector of $A$, with all entries of $v$ equal to a constant $k$. Naturally, $k \ne 0$. Let $\lambda$ be the eigenvalue of $A$ that has $v$ as an eigenvector. If $(b_1, b_2, \dots, b_n)$ is any row of $A$, then by the definition of eigenvalue and eigenvector, we have
$$kb_1+kb_2+\cdots +kb_n=\lambda k,$$
from which we conclude that $b_1+b_2+\cdots+b_n=\lambda$. It follows that each row sum of the matrix is equal to $\lambda$.
Conversely, suppose that all row sums of $A$ are equal to $\sigma$. Let $v$ be the vector with all entries equal to $1$. Then $Av$ is a vector with all entries equal to $\sigma$, which means that $v$ is an eigenvector of $A$ with eigenvalue $\sigma$.
Thus $A$ has an eigenvector with all entries equal if and only if all row sums of $A$ are equal.