Assuming that the Hadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \times n$ matrix that satisfies the given constraints is given by
$$\boxed{\mathrm A := \frac{1}{\sqrt n} \mathrm H_m^{\top} \mathrm S_n}$$
where
$\mathrm H_m \in \{\pm 1\}^{m \times m}$ is a Hadamard matrix. Thus, the $m$ rows of $\mathrm H_m$ are orthogonal, i.e., $$\mathrm H_m \mathrm H_m^{\top} = m \mathrm I_m$$
$\mathrm S_n$ is a thin $m \times n$ matrix whose $n$ columns are chosen arbitrarily from the $m$ columns of the $m \times m$ identity matrix. Thus, the $n$ columns of $\mathrm S_n$ are orthonormal, i.e.,
$$\mathrm S_n^{\top} \mathrm S_n = \mathrm I_n$$
Hence,
$$\mathrm A^{\top} \mathrm A = \frac{1}{n} \mathrm S_n^{\top} \mathrm H_m \mathrm H_m^{\top} \mathrm S_n = \frac{m}{n} \mathrm S_n^{\top} \mathrm S_n = \frac{m}{n} \mathrm I_n$$
as desired. Let $\mathrm e_k$ and $\mathrm h_k$ denote the $k$-th columns of $\mathrm I_m$ and $\mathrm H_m$, respectively. Hence,
$$\mathrm e_k^{\top} \mathrm A \mathrm A^{\top} \mathrm e_k = \| \mathrm A^{\top} \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm H_m \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm h_k \|_2^2 = \frac 1n \sum_{k=1}^n (\pm 1)^2 = \frac nn = 1$$
for all $k \in \{1,2,\dots,m\}$, as desired. Note that we used the fact that the entries of $\mathrm h_k$ are $\pm 1$.
If $m$ is a power of $2$, then $\mathrm H_m$ can be built using the Sylvester construction
$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_{2} = \begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}$$
if $m$ is not a power of $2$, we can use the Paley construction instead.