How to prove a multivariate r.v. does not follow the nonparanormal distribution? Background
You may find the definition of the non-paranormal distribution at the 2nd paragraph in p.2296 of this paper.
In short, $(X_1, \ldots, X_p)$ is non-paranormal if there exists a set of strictly increasing functions such that $(f_1(X_1), \ldots, f_p(X_p))$ is multivariate Gaussian.
Question
I'm trying to think of an example of a multivariate random variable that is not non-paranormal. For example, how do I check if the multivariate t-distribution (with some d.f.) follows the non-paranormal distribution or not? 
Any comments are very much welcome!
 A: First, a discrete distribution (or a distribution with atoms) cannot be transformed such to a multinormal, so assume an absolutely continuous distribution. Then, if $p=1$, we can always transform to normal. So assume $p>1$. Then, if the components are independent, again, we can use the univariate solution. So there must be some dependencies between the components. Then, this is essentially the same question as Can any probability distribution be transformed into a normal distribution?, which however do not give a very concrete example. 
One wonders why the authors of the paper you quoted choose the strange name non-paranormal. Their requirement is essentially that $X_1, \dotsc,X_p$ have a distribution with a Gaussian copula. So your question is really about giving examples of distributions with a non-gaussian copula (since copulas are unique.) That also makes it clear that there must be very many examples. That answers your question for the multivariate $t$, which defines a $t$-copula, not a Gaussian one. Some papers are this and this other.
One simple class of examples will be uniform distributions on polygons in the plane, with (at least some) edges not parallel to the $x$- or $y$-axes. To see why, use a diamond in the plane, a polygon with corners at $(0,1),(1,0),(0,-1), (-1,0)$. To have any chance at all of transforming to binormal, we must have that $\lim_{x \to 1} f_i(x) =+\infty$, and likewise $\lim_{x \to -1} f_i(x) =-\infty$, for if these limits where finite, we would get a finite range. But the image of the diamond cannot be all of the plane! so the distribution cannot be binormal. Note that here only the geometry plays a role, not the uniformity. 
A: I think the following example could do it : 
Let $X_1 \sim \mathcal{N}(0, 1)$ and $B$ be a binary variable which is equal to $1$ with probability $0.5$ and $-1$ with probability $0.5$. Define $X_2 = B\times X_1$. 
Then the marginal distributions of $X_1$ and $X_2$ are obviously $\mathcal{N}(0, 1)$ but $(X_1; X_2)$ is not a multivariate Gaussian. An easy way to see it is that $\mathrm{cov}(X_1, X_2) = 0$ whereas $X_1$ and $X_2$ are not independent, which would be impossible if $(X_1, X_2)$ was a multivariate Gaussian. 
The only increasing functions $f_1$, $f_2$ that with conserve $X_1$ and $X_2$ as a normal random variables are affine functions with positive coefficient, but they wont transform $(X_1, X_2)$ into a multivariate Gaussian (the argument stated before still applies). Thus there is no increasing functions $f_1$, $f_2$ such that $(f_1(X_1), f_2(X_2))$ is a multivariate Gaussian.
This is a bit of an artificial example, sorry for that. I would expect the multivariate $t$ distribution not to be non-paranormal, but derivation of that seems a bit tedious..
