The Kostka change of basis is dominance-unitriangular

Fix $d>0$, $\lambda ⊢d$, and rank $N\ge d$. Setting $\mu =\mathrm{\varnothing }$ in [F7] and using [F11] reduces the defining sum to $s_{\lambda \mathrm{/}\mathrm{\varnothing }}=s_{\lambda }$; [F8] therefore expresses the rank-$N$ specialization of $s_{\lambda }$ as the weight-monomial sum over semistandard $\lambda$-tableaux. For $\nu ⊢d$, the coefficient of $x_1^{{\nu }_1}{x}_2^{{\nu }_2}\cdots$ is $K_{\lambda \nu }$ by [F4]. By [F6], permuting variables preserves this polynomial, so every monomial in the orbit of $\nu$ has the same coefficient. Since $m_{\nu }$ is the sum of the distinct orbit monomials [F3], the coefficient of $m_{\nu }$ is $K_{\lambda \nu }$. The projection in [F2] identifies the rank-$N$ expansion with the stable one, giving $s_{\lambda }={\sum }_{\nu ⊢d}{K}_{\lambda \nu }{m}_{\nu }$. For $d=0$, this is $s_{\mathrm{\varnothing }}=m_{\mathrm{\varnothing }}=1$ with $K_{\mathrm{\varnothing },\mathrm{\varnothing }}=1$.

If a semistandard tableau $T$ has shape $\lambda$, each cell in row $i$ has a cell above it in every preceding row. Positivity and strict increase down columns force its entry to be at least $i$, so all entries at most $r$ lie in the first $r$ rows. A tableau of content $\mu$ has ${\sum }_{i=1}^r{\mu }_i$ entries at most $r$, hence ${\sum }_{i=1}^r{\mu }_i\le {\sum }_{i=1}^r{\lambda }_i$ for every $r\ge 1$. By [F5], $\lambda ⊵\mu$. If $\lambda$ does not dominate $\mu$, no such tableau exists and $K_{\lambda \mu }=0$.

Suppose $\lambda =\mu$. For each $r$, the first $r$ rows have exactly ${\sum }_{i=1}^r{\mu }_i$ cells, and all entries at most $r$ lie in those rows; the content supplies exactly that many such entries. Thus every cell in the first $r$ rows has entry at most $r$. Taking $r=i$ and using the lower bound at least $i$ from step 1.2 forces every cell in row $i$ to contain $i$. This filling is semistandard and unique, so $K_{\mu \mu }=1$. The empty shape has its unique empty tableau, giving the same conclusion for $d=0$.

By step 1.1 and [F10], $⟨h_{\mu },s_{\lambda }{⟩}_H={\sum }_{\nu ⊢d}{K}_{\lambda \nu }⟨h_{\mu },m_{\nu }{⟩}_H=K_{\lambda \mu }$. The form is symmetric: by [F11], writing $f={\sum }_{\alpha }{a}_{\alpha }{s}_{\alpha }$ and $g={\sum }_{\alpha }{b}_{\alpha }{s}_{\alpha }$ gives $⟨f,g{⟩}_H={\sum }_{\alpha }{a}_{\alpha }{b}_{\alpha }=⟨g,f{⟩}_H$. Therefore $⟨s_{\lambda },h_{\mu }{⟩}_H=K_{\lambda \mu }$; this symmetry follows from the proved orthonormal basis, not from an extra assumption on the defining pairing. Expand $h_{\mu }={\sum }_{\lambda ⊢d}{c}_{\lambda \mu }{s}_{\lambda }$ in the integral Schur basis [F11]. Pairing on the left with $s_{\kappa }$ gives $c_{\kappa \mu }=⟨s_{\kappa },h_{\mu }{⟩}_H=K_{\kappa \mu }$. Thus the stated expansion holds.

The set of partitions of $d$ is finite; order it by a linear extension of dominance from smaller to larger. By step 1.2, a nonzero off-diagonal entry $K_{\lambda \mu }$ can occur only when row label $\lambda$ follows column label $\mu$; by step 2.1 every diagonal entry is one. The entries are integers because they count finite sets [F4], so the matrix is lower unitriangular over $\mathbb{Z}$. For $d=0$ it is the one-by-one matrix $(1)$. Both families are integral bases [F9, F11], so this is their integral change-of-basis matrix.

In degree zero the empty tableau gives $h_{\mathrm{\varnothing }}=s_{\mathrm{\varnothing }}=1$, and in degree one the sole tableau gives $h_{(1)}=s_{(1)}$. Empty or impossible tableau sets give zero counts by definition; zero inputs pair to zero by bilinearity. Zero padding covers prefix sums beyond either partition's length. Each degree has finitely many partition labels and tableaux, so the finite order extension and expansions use no choice. No converse criterion is asserted.