Let $\mathfrak S_n$ be the symmetric group of permutations of $n$ letters and let $S = \sum_{\sigma\in\mathfrak S_n} \sigma$ be the symmetrization operator. Let $\Lambda_n^r$ be the vector space of symmetric polynomials in $n$ variables $\{x_1,\dots,x_n\}$ of (homogeneous) degree $r$.

Consider the map $$ \Psi: \Lambda_n^r\to\Lambda_n^r\\ f \mapsto S \cdot f(x_1, x_1 + x_2, \dots, x_1 + x_2 + \dots + x_n), $$ where the symmetric group acts by permuting variables.

Computational evidence suggests that, for $n\geq 1$, this map is bijective if and only if $r$ is even, or $n=1$, or $r=1$.

How could we prove this?

Notes:

Maybe it helps to generalize the question as follows. We could write the map as $$ f\mapsto S\cdot f(M \mathbf x) $$ where $\mathbf x = (x_1,\dots, x_n)^T$ and $M$ is an $n\times n$ matrix. In the special case above, $$ M = \left(\begin{matrix} 1 & 0 &\dots & 0\\ 1 & 1 &\ddots &\vdots\\ \vdots & &\ddots & 0\\ 1 & \dots & 1 & 1 \end{matrix}\right). $$ Which matrices $M$ make the map bijective? Note that bijectivity is preserved when permuting the columns or rows of $M$, because $f$ is symmetric and because we symmetrize $f(M\mathbf x)$.

Experimentally, for small $n$, there are many matrices that make $\Psi$ bijective. In fact, there are also many matrices that make $\Psi$ bijective if and only if $r$ is even.

In general, the map does not map characters to characters, for example, for $n=2$ and $r=6$.

Looking at the sequences of determinants, we have the following table, the first row corresponding to the case $n=1$, the first column corresponding to $r=1$: $$ \begin{array}{llllllll} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 3 & 2 & 0 & -2 & 0 & 4 & 0 & -4 \\ 12 & 40 & 0 & 560 & 0 & 26880 & 0 & -49674240 \\ 60 & 1200 & 0 & -1128960 & 0 & -682795008000 & 0 & 72912553997374586880 \\ 360 & 50400 & 0 & -23897825280 & 0 & -3663070894604943360000 & 0 & -1322279213561113068120489525050867712000 \\ \end{array} $$ Apparently, for $n=2$ we get, up to sign, $2^{\lfloor\frac{r}{6}\rfloor+1}$ for even $r$. It seems that the entries in the row for $n=3$ are nicer if we divide by the corresponding Bernoulli numbers. For $r=1$ we seem to get the order of the alternating group. Finally, for $r=2$ and $n>1$, the determinants seem to be $(n+2)!(n+1)!/72$.