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.