Let $\Lambda^d_n$ the space of symmetric polynomials in $n$ variables, with maximum 'partial degree' of each variable $d$. A basis for this space is the set of symmetrized monomials $m_\lambda$, where $\lambda$ is a partition with maximally $n$ parts, with each part $\leq d$.

Take $n= m N$ and $d = N$ (with $m>1$, and both $m$ and $N$ finite) and define the following specialization (or plethysm) $\mathcal{C}$, $$ \mathcal{C}: \Lambda^{N}_{m N} \rightarrow \Lambda^{mN}_{N} $$ that conflates the $m N$ variables to $N$ variables, via $x_{m(i-1)+j} \rightarrow y_i$, for $i=1,2,\ldots,N$ and $j=1,2,\ldots,m$. Specifically, for $p$ a symmetric polynomial in $\Lambda^N_{m N}$, we have $$ \mathcal{C} \bigl( p(x_1,\ldots,x_{mN}) \bigr) = p( \underbrace{y_1,\ldots,y_1}_{m}, \underbrace{y_2,\ldots,y_2}_{m},\ldots,\underbrace{y_{N},\ldots,y_{N}}_{m}) \ . $$

We strongly suspect that the map $\mathcal{C}$ as defined above is bijective, but could not find this or a similar result in the literature. Is it known that $\mathcal{C}$ is indeed bijective, and if so, how does one prove this?

Note that the restriction on the `partial degree' is essential for $\mathcal{C}$ to be injective. A simple example shows that $\mathcal{C}$ is not injective on, for instance, $\Lambda_2$, where $\Lambda_n$ is the space of symmetric polynomials in $n$ variables. Namely, take $p_2 (x_1,x_2) - 1/2 (p_1(x_1,x_2))^2 = 1/2 (x_1-x_2)^2$. It is clear that $\mathcal{C} \Bigl( p_2 (x_1,x_2) - 1/2 (p_1(x_1,x_2))^2 \Bigr) = 0$, showing that $\mathcal{C}$ is not injective on $\Lambda_2$.

We tried to find a pair of bases, for which the transition matrix becomes triangular, but did not succeed. Take for instance symmetrized monomials. They can simply be expressed in terms of power sums, for which the action of $\mathcal{C}$ is simple. Expressing the result back in terms of monomials (the power sums are in general not linearly independent), leads to the result that the transition matrix is not triangular, as can be shown by means of a simple example.

Take $m=2$ and $N=3$, and consider the polynomials with total degree 4. The basis for $\Lambda_6$ is $m_\lambda (x)$, with $\lambda \in \{ (4),(3,1),(2,2),(2,1,1),(1,1,1,1) \}$, while the basis for $\Lambda_3$ is given by $m_\lambda (y)$, with $\lambda \in \{ (4),(3,1),(2,2),(2,1,1)\}$. Note that for three variables, $m_{(1,1,1,1)} = 0$. The action of $\mathcal{C}$ is given, in these bases, by $$ \begin{pmatrix} 2 & 2 & 1 & 0 & 0\\ 0 & 4 & 0 & 4 & 0\\ 0 & 0 & 4 & 4 & 1\\ 0 & 0 & 0 & 8 & 4\\ \end{pmatrix} $$ Taking into account the `partial degree' restriction amounts to deleting the first column of this matrix. The resulting matrix has non-zero determinant, but this does not follow from a triangular structure.