Does there exist a non-identity matrix that preserves the ternary dot product?

Question

This is an variant on an older question.

Define the ternary dot product

\begin{align*} \mathbb{R}^n,\mathbb{R}^n,\mathbb{R}^n &\rightarrow \mathbb{R} \\ x,y,z &\mapsto x \bullet y \bullet z \end{align*}

as follows:

$$x \bullet y \bullet z = \sum_{i \in n} x_iy_i z_i$$

For example: $$(x_0,x_1) \bullet (y_0,y_1) \bullet (z_0,z_1) = x_0 y_0 z_0 + x_1 y_1 z_1$$

More generally, there's a $k$-ary dot product $$\overbrace{\mathbb{R}^n,\ldots,\mathbb{R}^n}^k \rightarrow \mathbb{R}$$ defined in the obvious way.

Questions.

Q0. $(k=3)$. Does there exist an $n \geq 0$ together with an $n \times n$ matrix $A$ distinct from $I_n$ satisfying the following identity? $$Ax \bullet Ay \bullet Az = x \bullet y \bullet z$$

Q1. $(k=4)$. Does there exist an $n \geq 0$ together with an $n \times n$ matrix $A$ distinct from $I_n$ and $-I_n$ satisfying the following identity? $$Aw \bullet Ax \bullet Ay \bullet Az = w \bullet x \bullet y \bullet z$$

Answer 1

Yes to both. For $n = 2$, let $A = \pmatrix{0 & 1\cr 1 & 0\cr}$. Or for general $n$, take any permutation matrix.