Let $A, B \in \mathbb{R}^{n \times n}$. The polyhedral cones of $A$ and $B$ are given by $$\mathcal{C}_A = \{ x \in \mathbb{R}^n : x = A \lambda, \mbox{ where } \lambda_i \ge 0 \mbox{ for all } i = 1, 2, \ldots n\}, $$ and $$\mathcal{C}_B = \{ x \in \mathbb{R}^n : x = B \lambda, \mbox{ where } \lambda_i \ge 0 \mbox{ for all } i = 1, 2, \ldots n\},$$ respectively. The polyhedral cones are all of the linear combinations of the columns of the corresponding matrix with non-negative weights.
Let $\pi_{i,j} : \mathbb{R}^n \rightarrow \mathbb{R}^2$ be the matrix which projects $\mathbb{R}^n$ onto its $i^{\text{th}}$ and $j^{\text{th}}$ coordinates.
Question: If $\pi_{i,j}( \mathcal{C}_A ) \subset \pi_{i,j} (\mathcal{C}_B)$ for every $i \neq j$, then is it necessarily true that $\mathcal{C}_A \subset \mathcal{C}_B$?
It is fairly easy to come up with a counter example for arbitrary sets in $\mathbb{R}^n$, for example these two ellipses in $\mathbb{R}^3$, but I have not been able to find a counter example when the sets are polyhedral cones. In any counterexample I can find, the intersection between the two sets is being "hidden" from the projection down to the coordinate planes. My intuition is that this is not possible for polyhedral cones. Can anyone prove that this is true or produce a counter example?
Edit: copper.hat found a counterexample where $\pi_{i,j}( \mathcal{C}_A ) = \pi_{i,j} (\mathcal{C}_B)$ for all $i,j$. Does a counterexample still exist for the statement
If $\pi_{i,j}( \mathcal{C}_A ) \subsetneq \pi_{i,j} (\mathcal{C}_B)$ for all $i,j$, then $\mathcal{C}_A \subset \mathcal{C}_B$?
Let $A=\begin{bmatrix} -2 & 2 & -2 \\ 1 & 1 & -2 \\ 1 & 1 & 1 \end{bmatrix}$, $B =\begin{bmatrix} -2 & 2 & 2 \\ 1 & 1 & -2 \\ 1 & 1 & 1 \end{bmatrix}$. Then the projections are all the same but the cones are different.