Let $\vec{x} \in \mathbb{R}^n$ be a fixed vector and suppose that we are given an isotropic random vector $\vec{a} = (a\_1, \dots, a\_n)^T$ in $\mathbb{R}^n$ (i.e., the covariance matrix of $\vec{a}$ is the identity matrix).


Are there some general conditions on $\vec x$ and the distribution of $\vec a$ such that the following holds?
There exist a (non-linear) mapping $P: \mathbb{R}^n \to \mathbb{R}^n$, depending on $\vec{x}$, statisfying the following two properties:


1. The entries of $P(\vec a)$ and $\langle \vec a, \vec x\rangle$ are independent.
2. The range of $Q := Id - P$ is one-dimensional, i.e., there exists some $\vec{x}' \in \mathbb{R}^n$ with $range(Q) \subset span\{\vec x'\}$.


**Edit:** One specification that I am particularly interested in looks as follows: Assume that the entries of $\vec a$ are deformed Gaussians, i.e., $a\_i = F(g\_i)$ with $g\_i \sim N(0,1)$ i.i.d. and $F$ continuous and invertible.


My intention is to generalize the well-known special case of independent Gaussians, $a\_i \sim N(0,1)$, where one can simply choose $P$ to be the orthogonal projection onto $\{\vec x\}^\perp$ and $\vec x' = \vec x$.
This works due to the rotation invariance of isotropic Gaussian random vectors and I wonder to what extend this can be relaxed. 



Independent Decomposition of a Random Vector
This will not work in general. Let's take $a$ as a vector with iid $\pm 1$ entries, with probability $1/2$ for each of $\pm 1$, and $x=(1,1,\ldots ,1)$ (normalize this if you prefer). Then if $\langle a, x\rangle = n$, we know for sure that $a=(1,1,\ldots , 1)$, so $P(a)$ would have to be constant almost surely if we want it independent of $\langle a, x\rangle$.


If you compare this with your example (normal distribution), then you see that this had the crucial extra property that the entries stay independent after changing coordinates by a rotation.


