I am wondering whether a certain instance of combinatorial reciprocity (in the sense of Stanley's classic paper "Combinatorial Reciprocity Theorems"), concerning symmetric functions, is known or follows from a more general result.
I use standard notion for symmetric functions as in e.g. Stanley's EC2.
For a symmetric function $f \in \Lambda$, and vector $\mathbf{x}=(x_1,\ldots,x_k)$ of fixed complex numbers $x_1,\ldots,x_k \in \mathbb{C}$, define $$\Phi(f,\mathbf{x};n) = f(\overbrace{x_1,\ldots,x_1}^{\textrm{$n$ times}},\overbrace{x_2,\ldots,x_2}^{\textrm{$n$ times}},\ldots,\overbrace{x_k,\ldots,x_k}^{\textrm{$n$ times}}).$$
It is clear that for power sum symmetric functions we have $\Phi(p_m,\mathbf{x};n) = n \cdot p_m(x_1,\ldots,x_k)$. Hence, since the $p_i$ generate $\Lambda$ as a ring, for any $f \in \Lambda$ we have that $\Phi(f,\mathbf{x};n)$ is a polynomial in $n$. So we can make sense of $\Phi(f,\mathbf{x};-n)$.
Proposition For any $f \in \Lambda$ of degree $m$, $\Phi(f,\mathbf{x};-n) = (-1)^{m} \cdot \Phi(\omega(f),\mathbf{x};n)$, where $\omega\colon \Lambda \to \Lambda$ is the canonical involution on the ring of symmetric functions swapping $e_n$ and $h_n$.
Proof: It suffices to check on a basis of $\Lambda$. For power sum symmetric functions we have $\Phi(p_{\lambda},\mathbf{x};n) = n^{\ell(\lambda)} p_{\lambda}(x_1,\ldots,x_k)$ and $\omega p_{\lambda} = (-1)^{|\lambda|-\ell(\lambda)} p_{\lambda}$, which together show that the proposition holds for the $p_{\lambda}$. $\square$
Using the standard non-intersecting paths interpretation of Schur functions, another, arguably more combinatorial proof of this proposition (showing $\Phi(s_{\lambda},\mathbf{x};-n)= (-1)^{|\lambda|}\Phi(s_{\lambda^t},\mathbf{x};n)$) can also be extracted from Gjergji Zaimi's beautiful answer to a previous MO question of mine, in which he establishes a general reciprocity theorem for non-intersecting paths in "growing" planar networks.
Question: Is this instance of combinatorial reciprocity known? Is it e.g. an example of Ehrhart-Macdonald reciprocity? Is there anything more interesting to be said about it?