Proposition 2III.1.P0251 of 76 in the corpus
The arrangement is the information, not the values.
Two images with identical pixel statistics can be a photograph and a jumble. Everything a vision architecture is built to exploit is in the arrangement, and the arrangement is exactly what a dense layer discards.
Demonstration
Take an image and apply a fixed permutation to its pixels. The mean, the variance, the histogram, every order-free statistic you can compute: unchanged. One of the two is a picture and the other is not.
That difference lives entirely in which value sits next to which, and it is the thing two priors are built to exploit.
Locality. Nearby pixels are related and distant ones usually are not. A convolution reads a small window, so it assumes this by construction.
Translation equivariance. The same object means the same thing wherever it sits. A convolution slides one kernel everywhere, so it assumes this too.
Neither is learned. Both are architectural, imposed by whoever chose the layer, and both are true of photographs, which is why they were worth imposing.
Now the sharp version of the claim. A fully connected layer on flattened pixels is invariant to a fixed permutation of its inputs in the following sense: apply any permutation to every image in the dataset and permute the first weight matrix’s columns the same way, and you have the identical model. It cannot prefer the un-permuted world, because nothing in its structure references adjacency.
So the shuffled image is not merely harder for a dense network. It is, before training, indistinguishable.
Corollary
This is why the flatten in a classifier goes at the end. Flattening first throws away the only structure the architecture was going to exploit, and every subsequent layer has to rediscover from data what a 3 × 3 kernel is given for free.