Can $f(X)\times E+g(X)$ be a Gaussian if $E$ is a Gaussian for non-constant $f$ and $g$? It is well-known that $aE+b$ preserves the normality of $E$ when $a$ and $b$ are constants. However it is unclear whether this can happen when said constants become (possibly) dependent random variables.
So for simplicity, let $E\sim N(0,1)$ and $X$ is a continuous random variable independent of $E$, I want to know if there is some non-constant functions $f(x)$ and $g(x)$ such that $f(X)\times E+g(X)\sim N(0,1)$?
 A: Your distribution is now that of two variables, $P(X,E)$, but for every value of $x$ the distribution of $E$, that is $P(E|X)$, is still Gaussian/Normal. It would be trickier, if $E$ and $X$ were correlated.
A: Generally the answer is no, loosely understanding "non-constant" also bans trivialities involving sets of probability zero.  But when $X$ also has a standard Normal distribution and $|\rho|\lt 1,$ notice that
$$Y = \rho\operatorname{sgn}(X) E + \sqrt{1-\rho^2}\,X$$
also has a standard Normal distribution.  The coefficient function $f(x) = \rho\operatorname{sgn}(x)$ and $g(x) = \sqrt{1-\rho^2}\,x$ are almost surely non-constant.
This claim can be demonstrated by noting (i) the transformation $(x,e)\to (x,\operatorname{sgn}(x)e)$ is almost surely differentiable and one-to-one; (ii) where it is differentiable, its Jacobian is $1;$ and (iii) the bivariate Normal density is unchanged by negating $e.$
BTW, an analogous construction applies to any bivariate Normal variable $(X,E)$ with zero mean.
A: Consider $f(x) = \pm 1/2$, each with probability $1/2$.  Then $f(x)\cdot E \sim N(\mu=0, \sigma^2=1/4)$.  Now add $g(x) \sim N(\mu=0,\sigma^2=3/4)$, and we have the result that $f(x)\cdot E + g(x) \sim N(0,1)$.
For examples of functions, let $X \sim N(\mu=0, \sigma^2=3/4)$, $g(x) = x$, and $f(x) = 0.5\cdot \mathrm{sign}(x)$.
An alternative:  Let $x \sim \mathrm{U}(0,1)$ and $f(x) = \mathrm{floor}(x+1/2)$.  Then $f(x) \cdot E$ is a mixture of a standard Gaussian with probability $50\%$ and a point mass at zero with probability $50\%$. Now, let $g(x) = 0$ if $x > 1/2$ and $\Phi^{-1}(2x)$ (the inverse Gaussian CDF) otherwise.  The resulting expression is also distributed standard Gaussian; when $f(x)\cdot E = 0$, we are drawing from a standard Gaussian in the function $g(x)$, and when $f(x) \cdot E \sim N(0,1)$, $g(x) = 0$, so the sum is still a standard Gaussian.
