From probability distribution in $\mathbb{R}^3$ to probability distribution in $\mathbb{R}^4$ I am working on a research paper where I need to investigate conditions for the existence of probability distributions satisfying certain characteristics. I have already asked a related question (here), whose answers allowed me to frame better in my mind the problems I'm facing. In what follows, I report a close, although different, question. I will highlight the key differences below.
Consider a $6\times 1$ random vector
$$
\eta\equiv (\eta_1,\eta_2,..., \eta_6)
$$
satisfying the following property (hereafter, called Property 1):
Property 1: $$
 \begin{pmatrix}
\eta_1\\
\eta_2\\
\eta_3
\end{pmatrix}  \sim  \begin{pmatrix}
\eta_4\\
-\eta_2\\
\eta_5
\end{pmatrix}   \sim  \begin{pmatrix}
\eta_6\\
-\eta_3\\
-\eta_5
\end{pmatrix}  \sim  \begin{pmatrix}
-\eta_1\\
-\eta_4\\
-\eta_6
\end{pmatrix}  \sim G
$$
where "$\sim$" denotes "distributed as" and $G$ is an absolutely continuous distribution with full support on $\mathbb{R}^3$.

Question A: Let $\mathcal{G}$ denote the family of absolutely continuous distribution with full support on $\mathbb{R}^3$ and whose marginals are symmetric around zero and identical. For each $G\in \mathcal{G}$, does there exists a vector $\eta$ satisfying Property 1?

Question B: Let $\epsilon$ be a $4\times 1$ random vector
$$
\epsilon\equiv \begin{pmatrix}
\epsilon_1\\
\epsilon_2\\
\epsilon_3\\
\epsilon_0\\
\end{pmatrix} $$
For each $(G,\eta)$ satisfying Property 1, does there exist $\epsilon$ satisfying Property 2 described below?
Property 2:
$$ 
\begin{pmatrix}
1 & 0 & 0 & -1\\
1 & -1 & 0 & 0\\
1 & 0 & -1 & 0\\
0 & 1 & 0 & -1\\
0 & 1 & -1 & 0\\
0 & 0 & 1 & -1\\
\end{pmatrix}*\epsilon=\begin{pmatrix}
\eta_1\\
\eta_2\\
\eta_3\\
\eta_4\\
\eta_5\\
\eta_6
\end{pmatrix}
$$
and the distribution $F$ of $\epsilon$ is  absolutely continuous  with full support on $\mathbb{R}^4$?

My thoughts:
I believe that the answer to Question A is "Yes": any  distribution in $\mathcal{G}$ satisfies Property 1. Certainly, there exist distributions outside $\mathcal{G}$ that can also  satisfy Property 1.
I believe that the answer to Question B is "Yes" as well. However, I'm not 100% sure and I would appreciate your help. The answers here suggest that: if $F$ is  absolutely continuous  with full support on $\mathbb{R}^4$, then $G$ is  absolutely continuous  with full support on $\mathbb{R}^3$. Here, however, I'm asking something different: if $G$ is  absolutely continuous  with full support on $\mathbb{R}^3$, can we always find a distribution $F$ for $\epsilon$ that is absolutely continuous  with full support on $\mathbb{R}^4$?
 A: I think the answer to both questions is negative.
Question A.
You ask whether for all $G\in\mathcal{G}$, there exist a random vector $\eta=(\eta_i)_{1\le i\le 6}$ such that they satisfy Condition 1 together.
If $G$ were not assumed to be fully supported, the answer would be easily seen to be negative: taking $G$ supported on $\{(x,y,z)\in\mathbb{R}^3 \mid x=y\}$ would force $\eta_1=\eta_2$, $\eta_4=-\eta_2$ and $-\eta_1=-\eta_4$ almost everywhere, which are incompatible.
But you can start from there and change $G$ slightly to be fully supported: simply start with any distribution supported on $\{(x,y,z)\in\mathbb{R}^3 \mid x=y\}$ with equal and centrally symmetric marginals; assume further that the marginals give a mass less than $1/5$ to $[-1,1]$ (added in edit: previously was $1/4$, but we need some room because of the perturbation). Define $G$ as a convolution of that distribution with a Gaussian $\sim\mathcal{N}(0,\varepsilon)$ for some small positive $\varepsilon$. Taking $\varepsilon$ small enough, you can ensure $G(\{(x,y,z)\in\mathbb{R}^3 \mid \lvert x-y\rvert>1/4\})<1/4$ and that any marginal of $G$ give a mass less than $1/5$ to $[-1,1]$. If a random vector $\eta$ were to satisfy condition 1 with $G$, you would have
\begin{align*}
\mathbb{P}(\lvert \eta_4\rvert\le 1) &<1/4 \\
\mathbb{P}(\lvert \eta_1-\eta_2\rvert>1/4) &<1/4 \\
\mathbb{P}(\lvert \eta_4+\eta_2\rvert>1/4) &<1/4 \\
\mathbb{P}(\lvert \eta_1-\eta_4\rvert>1/4) &<1/4 \\
\end{align*}
With positive probability, we would thus have $\lvert \eta_1-\eta_2\rvert\le 1/4$, $\lvert \eta_4+\eta_2\rvert\le 1/4$, $\lvert \eta_1-\eta_4\rvert \le1/4$ and $\lvert \eta_4\rvert > 1$. Now, this is impossible since
$$\lvert \eta_4+\eta_4\rvert \le \lvert \eta_4-\eta_1\rvert + \lvert\eta_1-\eta_2\rvert+\lvert\eta_2+\eta_4\rvert. $$
(I took more room than needed, but that does the trick.)
Question B. (added in edit)
Simply take $\eta$ a normal vector $\sim\mathcal{N}(0,I_6)$. Then condition 1 holds with $G$ a normal distribution, but for any $\epsilon$ its image under the matrix is contained in its image vector space, which has dimension $4$ at most since it is a $6\times 4$ matrix. This cannot have a fully supported law, hence cannot equal $\eta$.
