= Solution
Since $S^\mu\subset M^\mu$, Question 2 shows that every <composition factor> of $S^\mu$ is $D^\lambda$ for a <regular partition> $\lambda\unrhd\mu$. If $\mu$ is regular, its radical quotient $D^\mu$ occurs once: it occurs as a quotient, and its multiplicity is bounded above by the multiplicity one in $M^\mu$. All remaining factors have strictly dominating labels. A nonregular row has only strictly dominating regular labels.
Define $d_{\mu\lambda}=[S^\mu:D^\lambda]$. Order the row labels, all <partitions of an integer> of $n$, and the column labels, the regular <partitions of an integer>, by the same total order refining decreasing dominance. Then
$$
\boxed{d_{\mu\lambda}=0\text{ unless }\lambda\unrhd\mu,
\qquad d_{\lambda\lambda}=1.}
$$
The matrix is generally rectangular. Its submatrix on regular rows is lower unitriangular in this decreasing convention; placing least dominant shapes first would transpose the visual triangular convention, not the indices.
This is the <decomposition matrix (modular representation theory)>, not just an abstract <composition factor> table. The <standard polytabloid basis> makes $S^\mu_{\mathbb Z}$ a free invariant integral lattice whose extension to <characteristic zero> is the ordinary <irreducible> $S^\mu$. Reducing this same lattice modulo $p$ gives exactly the modular $S^\mu$. On <p-regular elements>, reduction of its ordinary <character> is the sum of the <Brauer characters> of its <composition factors>, with coefficients $d_{\mu\lambda}$. These are the defining decomposition numbers; the resulting class in the <modular representation ring> is independent of the chosen invariant lattice.
For $S_3$, take rows $(3),(2,1),(1,1,1)$ in that order. In <characteristic zero>, or $p>3$, all three <Specht modules> are <irreducible> by <Maschke's theorem>, and
$$
\boxed{M=I_3.}
$$
For $p=2$ the regular labels are $(3),(2,1)$ and
$$
\boxed{M=\begin{pmatrix}1&0\\0&1\\1&0\end{pmatrix}.}
$$
Indeed the <sign representation> equals the <trivial representation>. The two-dimensional $S^{(2,1)}$ is the sum-zero subspace of the three-point <permutation module>. Over an algebraic closure, a $3$-cycle has two distinct nontrivial third-root <eigenvalues> there, and a <transposition> exchanges their lines. Neither line is invariant under $S_3$, proving <absolute irreducibility>.
For $p=3$ the regular labels are $(3),(2,1)$ and
$$
\boxed{M=\begin{pmatrix}1&0\\1&1\\0&1\end{pmatrix}.}
$$
In the same sum-zero space the vector $(1,1,1)$ now spans an invariant trivial line. Its one-dimensional quotient is sign: the <determinant> of the sum-zero representation is sign, and the <determinant> of the invariant line is trivial. Therefore $S^{(2,1)}$ has one trivial and one sign factor, with sign its simple <head> $D^{(2,1)}$. The bottom <Specht module> is itself sign. This proves every entry of the three possible matrices.
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