"All 8x8 s-boxes created using Galois Field inversion plus an Affine Transform have the same non-linearity, as the Affine Transform does not change the linear or differential properties" is what this answer claims.
Is there a proof of this statement?
I am unable to see how this is true. I tried to calculate the Walsh and autocorrelation spectrum but was unable to see how they are equal.
To be precise, different affine transformations following the Galois field inversion should give the same Walsh spectrum up to a $\pm$ sign, in terms of how many times each value occurs. Since an affine transformation is a linear transformation plus a constant vector addition, this is not surprising.
Don't forget linear cryptanalysis measures distance to the unbiased case (prob. 1/2) so signs of Hadamard coefficients switching between positive and negative is allowed. This is how come we can ignore the sum of the non targeted key bits modulo 2 during Linear cryptanalysis, since all they would do is switch the sign of the relevant coefficient.
Let $$L_{a,b} := \sum_{x \in V_n} (-1)^{a \cdot x \oplus b \cdot S(x)}$$
where $V_n$ is the n dimensional binary vector space. Let $A x+c$ be an affine map where the linear part $x\mapsto Ax$ is full rank and thus invertible. It is then a simple matter of algebra to prove the result: $$L_{a,b}' := \sum_{x \in V_n} (-1)^{a \cdot x \oplus b \cdot (A \cdot S(x)\oplus c)}$$ by a change of basis.
Even better, in "The Design of Rijndael" by Daemen and Rijmen, (See here, Appendix A.1 onwards, provided for personal research use only) there is a coordinate free approach using trace functions on the finite field to show this.
Similar comments apply for the correlation spectrum.
