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 $n\geq 2$, this map is
bijective *if and only if* $r$ is even.

**How could we prove this?**

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**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 $r=6$.