Is it possible for $X$ and $Y$ to be marginally normally distributed and have $ E[Y|X] $ be a nonlinear function of $X$? Is this at all possible?  What is the intuition for this?  
 A: This is possible, here's one example that comes to mind.
Let $X ~ \sim \mathcal{N}(0,1)$ and let $Y$ have a truncated standard normal distribution on $[0, \infty)$ if $X \geq 0$ and on $(-\infty, 0)$ if $X < 0$. Then the marginal distributions of $X$ and $Y$ are both standard normal, but $\mathbb{E}(Y|X)$ is a non-linear function of $X$.
In particular, if $X \geq 0$ then $Y$ has a half-normal distribution so has mean $\sqrt{\frac{2}{\pi}}$, and by symmetry if $X < 0$ then $Y$ has mean $-\sqrt{\frac{2}{\pi}}$. Overall $\mathbb{E}(Y|X)=\sqrt{\frac{2}{\pi}}$ for $X \geq 0$ and $-\sqrt{\frac{2}{\pi}}$ for $X < 0$ so is indeed a non-linear function of $X$.
My personal intuition for this choice was to think about "chunks of probability" of $X$ - then on each chunk, what conditional distribution could I set for $Y$ that would ensure both that (i) the marginal distribution of $Y$ is normal, (ii) the conditional mean of $Y$ is not a linear function $X$. The latter can be achieved easily if a discrete chunk of $X$'s probability distribution is taken and the same conditional distribution of $Y$ is used for all $X$ in that range. This means that $\mathbb{E}(Y|X)$ is the same for all $X$ in that interval, i.e. is just a constant function of $X$. So long as it is a different constant function of $X$ over a different interval of $X$, then the conditional mean  must be a non-linear function of $X$ overall. 
I went for the easiest case of splitting the distribution of $X$ up into two parts with probability of $\frac{1}{2}$ each. After that, it was clear that the half-normal distribution would be a good choice for the conditional distribution of $Y$ - it's just a matter of gluing two sides back together to get the entire normal distribution as the marginal. If I'd wanted to something fancier, I could have cut $X$ up into many such chunks, possibly irregular, and selected appropriate truncated normals for the conditional distributions of $Y$ over each chunk. The question asks for the underlying intuition, and this is the thought process that resuilted in my solution, but if you want to generalise more widely you are best to think about the copula.
