Let $A \in \mathbb{R}^{n \times m}$ ($n>m$). Does exist a matrix $N \in \mathbb{R}^{n \times m}$ such that $N^TN=I_m$ and $A^TN=(A^TN)^T$. I easily found examples of matrices $N$ with $N^TN=I_m$ or $A^TN=(A^TN)^T$ but none working for both.
I also tried to set equations for $n=3$ and $m=2$. For $A=\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \\ \end{bmatrix}$ and $N=\begin{bmatrix} n_1 & m_1 \\ n_2 & m_2 \\ n_3 & m_3 \\ \end{bmatrix}$, we need to satisfy: \begin{align*} \sum_{i=1}^3 n_i^2 &= 1 \\ \sum_{i=1}^3 m_i^2 &= 1 \\ \sum_{i=1}^3 n_im_i &=0 \\ \sum_{i=1}^3 a_i n_i &= \sum_{i=1}^3b_i m_i \end{align*} but I can not conclude something since we have a set of non-linear equations.
Yes. Let $A=USV^T$ be a compact/economic singular value decomposition over $\mathbb R$, so that $U$ has the same size as $A$ and it has orthonormal columns, $S$ is a square nonnegative diagonal matrix, and $V$ is an orthogonal matrix. Now take $N=UV^T$. Then $N$ has orthonormal columns and $A^TN=VSV^T$ is symmetric.