Row sum and column sum of a square matrix

It becomes convenient to introduce complex numbers here.

$A^T A = AA^T = AA^* = A^*A$

so $A$ commutes with its conjugate transpose given by $A^* = A^T$, which means $A$ is normal and thus unitarily diagonalizable. So, selecting unitary matrix $U$, this implies

$U^{-1} A U =U^* A U = D$
or equivalently
$U^* A = DU^*$

your problem about row sums being equal to one implies an eigenvector equation, i.e. $A\mathbf 1 = \mathbf 1$. Your problem then wants you to confirm that $\mathbf 1^T A = \mathbf 1^T$. But this is implied by the above unitary diagonalization.

You may select (/assume WLOG) that $\mathbf u_1 = \mathbf 1$. I.e. the diagonal matrix $D$ is given by
$D = \pmatrix{1 & \mathbf 0^*\\\mathbf0& D_0}$

putting this all together
$\pmatrix{\lambda_1 \mathbf u_1^* \\*} = \pmatrix{ \mathbf 1^T \\*} = \pmatrix{1 & \mathbf 0^*\\\mathbf0& D_0}U^* = DU^* = U^* A$

thus
$\lambda_1 \mathbf u_1^* = 1 \cdot \mathbf 1^T = \mathbf 1^T = \mathbf 1^T A$
i.e. $\mathbf 1$ is a left eigenvector for $A$

Since $A^TA = AA^T$ and $AJ = J$ then \begin{align} A(A^TJ)= A^T(AJ) = A^TJ \end{align} i.e. $A^TJ$ is an eigenvector of $A$ with eigenvalue $1$. This also means $A^TJ-J$ is an eigenvector of $A$ with eigenvalue $1$ or equivalently, \begin{align} A^TJ-J \in E(1)= \text{Nul}(A-I) = \text{Row}(A-I)^\perp \end{align}

But, we also know that $A^TJ-J \in \text{Row}(A-I)=\text{Col}(A^T-I)$. Hence it follows $A^TJ-J=0$ which is the desired conclusion.