Let $A$ be a complex Hermitian matrix of order $n$ such that $A\mathbf{1}_n=k\mathbf{1}_n,$ where $k$ is some real constant and $\mathbf{1}_n$ is the vector of length $n$ and whose each component is $1$.
I have observed that if $x$ ($\perp\mathbf{1}_n$) is an eigenvector of $A$ affording eigenvalue $\lambda$, then there exists some other $\mu$ such that $\bar x$ (the conjugate vector of $x$) is also an eigenvector of $A$ affording eigenvalue $\mu$, that is, if $$Ax=\lambda x,$$ then there exists $\mu$ satisfying $$A\bar x=\mu\bar x.$$
But I am not sure about it and not even getting any counter example to it.
So my question is if it is true, then how to prove it and further how to establish a relationship between $\lambda$ and $\mu$.