Is Shift and Scalar invariant same as independent under linear transforms? I saw "shift and scale invariant" terms for the first time, and I'm wondering what's their meaning? in other word: Is Shift and Scalar invariant same as invariant under linear transforms?
thanks.
 A: The paper you linked answers this question:

In contrast, $T_n$ is not invariant under orthogonal transformations, but it is invariant under location shifts and scalar transformations.

Orthogonal transformations are linear, so it would seem the answer is no.
Location shifts and scalar transformation seem to have a domain specific definitions that I've not encountered before.  From the same paper

Here, the location shifts and scalar transformations mean $X_{ij} \mapsto B X_{ij} + c$ for $i=1,2, \ldots$, $j=1,\ldots,n_i$, where $c$ is a constant vector, $B=\text{diag}(b_{21},...,b_{2p})$, and $b_{21}, ..., b_{2p}$ are non-zero constants.

They don't offer a definition of constant vector, but it seems they must mean a vector all of who's components are equal.  You'd have to read in detail to be sure.  As for scalar transformation, generally I'd expect all the diagonal entries to be equal for that.  Quite confusing use of terminology here.
A: According the the terminology in the paper, the scalar transformations (from Proposition 1) only allow full rank diagonal matrix products. Hence, as the multiplication by $0$ is a linear operation, you have your counter-example.
Following a constructive exchange with @Matthew Drury, I have found the paper Finite Groups Represented by Special Matrices, G. A. Miller, Transactions of the American Mathematical Society, 1916, which defines:

A square matrix in which each of the elements, with a possible exception of those of the principal diagonal, is equal to zero is called a quasi-scalar matrix

The transformations in the paper are abusively called linear. A better naming could be: "positive quasi-scalar" for the matrix part.
