Necessary and sufficient conditions for left and right eigenvectors to be equal

I am answering the question about the general case.

Let $A=B+C$ denote the decomposition of $A$ into hermitian and antihermitian part. The assumption that $u$ and $u^*$ are (right and left)eigenvectors with respect to $\lambda$ is then equivalent to the equations $$Bu+Cu=\lambda u,~Bu-Cu=\lambda^*u.$$ These equations are equivalent to $$Bu=\Re(\lambda)u,~Cu=i\Im(\lambda)u,$$ where $\Re(\lambda)$ and $\Im(\lambda)$ are the real and imaginary part. In other words, $u$ is an eigenvector of both $B$ and $C$.

I somewhat doubt that there is a more explicit expression in the entrys of $A$ (in the general case) because the whole situation is unchanged when undergoing a unitary similarity transformation.

First, notice that if $A$ is irreducible with non-negatives entries then $A^t$ is also irreducible. The eigenspaces associated to the Perron-Frobenius eigenvalue, $\lambda$, of $A$ and $A^t$ are one dimensional, by Perron-Frobenius theorem. Let $B=A-\lambda Id$ and $k$ its order.

Let $\mathrm{adj}(C)$ be the adjugate of $C$, i.e., $C\ \mathrm{adj}(C)=\det(C)Id$.

Since $B$ and $B^t$ have rank $k-1$ then $\mathrm{adj}(B)$ and $\mathrm{adj}(B^t)$ have rank 1. Notice that $$B\ \mathrm{adj}(B)=\det(B)Id=0_{k\times k}.$$

Therefore, the non-null columns of $\mathrm{adj}(B)$ are multiples of the Perron-Frobenius eigenvector of $A$ associated to $\lambda$. Analogously, the non-null columns of $\mathrm{adj}(B^t)$ are multiples of the Perron-Frobenius eigenvector of $A^t$ associated to $\lambda$

Thus, $A,A^t$ have the same Perron-Frobenius eigenvector if and only if $$B^t\mathrm{adj}(B)=0.$$