Question: Is it possible to have a matrix $\mathbf B \in \mathbb R^{m\times n}$ such that it satisfies:
$$\mathbf B^T\cdot\mathbf B = \mathbf I\_n$$

Where $\mathbf I\_n$ is the $n\times n$ identity matrix.

Or in other words, is it possible to have a rectangular matrix such that it's transpose is it's left inverse?

If so, what kind of matrix is this? How could you go about constructing such a matrix?

Answer: 
You can construct matrices with this property as follows: assume $m\geq n$ and endow $\Bbb R^m$ with the standard euclidean inner product.
Pick vectors ${\bf v}\_1,...,{\bf v}\_n$ such that ${\bf v\_i}\cdot{\bf v\_j}=\delta\_i^j$ where $\delta$ is the Kronecker symbol.

Then the matrix $B$ whose columns are the vectors ${\bf v}\_1,...,{\bf v}\_n$ (in any coordinate system) has the desired property.
Here's an example: $B = \begin{pmatrix}1 \\ 0 \end{pmatrix}$ then $B^T\cdot B=1$. For something less simple, you can consider $B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$.