From consideration of a wider problem I get the following multiple matrixproduct and intuitively I assume, that matrixproduct cannot become the unit-matrix.
Let the example matrices be the following, but may be this can already be answered in general.
$$ A = \begin{bmatrix}
0 & -\sqrt{0.5}& \sqrt{0.5} \\
-\sqrt{0.5} & 1/2 & 1/2 \\
\sqrt{0.5}& 1/2 & 1/2 \\
\end{bmatrix} \qquad \qquad \text{ Note: } A=A^{-1}=A^t
$$
$$ D = \text{diagonal} \left(\begin{bmatrix}
1 & ß^{-1} & ß \end{bmatrix} \right)
\qquad \qquad \text{ where } ß = 3 + 2 \sqrt 2
$$
Q: I have now the fourfold parametrizable matrixproduct, and ask, whether -for any combinations of $\; k_j \in \mathbb Z \backslash 0 \;$ - I can get equality: $$ [x,y,z] \cdot A \cdot D^{k_1} \cdot A \cdot D^{k_3} \overset?= [x,y,z] \cdot D^{k_0} \cdot A \cdot D^{k_2} \cdot A \tag 1 $$
I think it is also equivalent to the question involving the matrices alone (without the leading row-dotmultiplication by vector $[x,y,z]$ ):
Q: can occur equality in the equation $$ A \cdot D^{k_1} \cdot A \cdot D^{k_3} \cdot A \cdot D^{-k_2} \cdot A \cdot D^{-k_0} \overset?= I \tag 2 $$
This latter form might have a general, elementary answer, but I'm unsure of that, and also, whether the removal of the row-dotproduct doesn't change the characteristic of possible equality in(1).
update: I see, that $B_k = A \cdot D^k \cdot A $ is symmetric. Then from (1) I take $$ B_{k_1} \cdot D^{k_3} \overset?= D^{k_0} \cdot B_{k_2} \tag 3 $$ On the lhs I've the columnscaling and on the rhs I've a rowscaling, so even more I think there should no (nontrivial:$k_j \ne 0$) solution. Is this observation sufficient?
(Impossibility of a solution for (1) seems to allow a sought disproof in my wider problem)