Why is the following statement false 

if an $n \times p$ matrix $U$ has orthonormal columns, then $UU^T\mathbf{x} = \mathbf{x}$ for all $\mathbf{x}$ in $\mathbf{R}^n$


So this is my understanding, 
if $U$ has orthonormal columns then $U^TU = I$, this is by Theorem A.
Thus $U$ is the inverse of $U^T$ and vice versa, thus $UU^T = U^TU = I$
Therefore $UU^T\mathbf{x} = (UU^T)\mathbf{x} = (I)\mathbf{x} = \mathbf{x}$
NOTE :
Theorem A : $\quad$
An $m \times n$ matrix $U$ has orthonormal columns if and only if $U^TU = I$

$U$ is not the inverse of $U^{T}$, since $U$ is not a square matrix. That's the mistake in your argument. 
