Consider the following integral representation of $\zeta(3)$:
$$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{1}{1-xyz}dxdydz$$
The integrand is a rational function of $d$ variables and the domain of integration $[0,1]^d$ with $d=3$. Now consider the group $H_{d}$ of permutations of the variables leaving the integrand invariant. Here it is $S_{3}$. Consider now the group $G_{d}$ of isometries of the domain of integration. This group is the isometry group of the cube $[0,1]^3$, hence $S_{4}\times C_{2}$, which admits ${S}_{3}$ as a subgroup.
Hence my question:
Is the degree $d(p)$ of a period $p$ expressed as an integral $I_{p}$ over $[0,1]^{d(p)}$ the smallest $d$ such that $H_{d}$, the group of permutations of the $d$ variables of integration preserving the expression of the integrand of $I_{p}$, is a subgroup of $G_{d}$, the isometry group of the domain of integration $[0,1]^d$?
If yes can we deduce that $d(\zeta(3))=3>1$ hence the transcendence of $\zeta(3)$?
That way we may define the "Kontsevitch-Zagier symmetry group" of a period $p$ of degree $d$ as $G_{d}/H_{d}$, and make an analogy between this group and the Galois group of a degree $d$ polynomial with rational coefficients, as the group of automorphisms of $\mathbb{Q}(x_{1},\cdots,x_{d})$ preserving the rational fraction that the integrand of $I_{p}$ is.