 Solution

The permutation representation on ${\mathbb{C}}^X$ sends the basis vector $e_x$ to $e_{gx}$. Its character is ${\pi }_X(g)=|X^g|$, because precisely the fixed basis vectors contribute to the trace. Burnside lemma says that the number of orbits is $|G{|}^{-1}{\sum }_g|X^g|$. To prove it, double-count $(g,x)$ with $gx=x$: the count is ${\sum }_g|X^g|={\sum }_x|G_x|$. Each orbit contributes $|Gx||G_x|=|G|$ by the orbit-stabilizer theorem.

A pair is fixed under the diagonal action exactly when both entries are fixed. Since permutation characters are real,by Burnside lemma on the product.

The general linear group ${\mathrm{GL}}_2(q)$ acts on the $q+1$ lines of ${\mathbb{F}}_q^2$. It is doubly transitive: choosing nonzero representatives of any two distinct lines gives a basis, and a linear map can send that basis to representatives of any other ordered pair. Thus there are two orbits on ordered pairs, the diagonal and its complement. The character inner product of this permutation character with itself is two, while the multiplicity of the trivial character is one by transitivity. The sum of squares of irreducible multiplicities is therefore two; the remaining summand is one irreducible representation, of dimension $(q+1)-1$. Its dimension is $\boxed{q}$.

For $S_n$ acting on two-element subsets, an ordered pair of subsets is classified by intersection size $0,1,2$. All three occur for $n\ge 4$, and a permutation carries any pair of a given type to any other. Hence $⟨{\pi }_Z,{\pi }_Z⟩=3$. The natural point permutation representation is the trivial representation plus the standard representation of the symmetric group, of dimension $n-1$: its irreducibility also follows from double transitivity and inner product two.

Embed this point representation into ${\mathbb{C}}^Z$ by $e_i\mapsto {\sum }_{j\ne i}{e}_{\{i,j\}}$. The map is equivariant, and its kernel consists of vectors with $c_i+c_j=0$ for every distinct $i,j$. Three distinct indices force every $c_i=0$, so it is injective for $n\ge 3$. Thus the trivial and standard representations each occur in ${\mathbb{C}}^Z$. The norm three leaves exactly one further irreducible, with multiplicity one. It is the orthogonal complement of that embedded subspace, equivalently the vectors $a_{ij}=a_{ji}$ satisfying ${\sum }_{j\ne i}{a}_{ij}=0$ for every $i$. Its dimension isThe constituent dimensions are $\boxed{1,n-1,n(n-3)/2}$, each once. For $n=3$, complementation identifies two-element subsets with points, so there are only the trivial and 2-dimensional standard constituents.