= Solution
Use the full definition of the <special unitary group>: $A^\dagger A=I$ and $\det A=1$. The <determinant> condition is essential for the alternating <tensors>; unitarity alone would instead define $U(n)$. Assume $n\geq2$ when discussing proper <invariant subspaces>.
Let $V=\mathbb C^n$ be the <fundamental representation>. A $(j,k)$ <tensor> belongs to $V^{\otimes j}\otimes(V^*)^{\otimes k}$. Its <SU(n) tensor transformation> is
$$
\boxed{T'{}^{\alpha_1\cdots\alpha_j}_{\beta_1\cdots\beta_k}
=A^{\alpha_1}{}_{\gamma_1}\cdots A^{\alpha_j}{}_{\gamma_j}
(A^{-1})^{\delta_1}{}_{\beta_1}\cdots(A^{-1})^{\delta_k}{}_{\beta_k}
T^{\gamma_1\cdots\gamma_j}_{\delta_1\cdots\delta_k}.}
$$
Since $(A^{-1})^\delta{}_{\beta}=\overline{A^\beta{}_{\delta}}$, <complex conjugation> exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline T$ is a $(k,j)$ <tensor>.
The <invariant tensors> follow directly. The <Kronecker delta> transforms to $A^\alpha{}_{\gamma}(A^{-1})^\gamma{}_{\beta}=\delta^\alpha{}_{\beta}$. For the alternating <tensors>, the <determinant> expansion gives
$$
A^{\alpha_1}{}_{\gamma_1}\cdots A^{\alpha_n}{}_{\gamma_n}\epsilon^{\gamma_1\cdots\gamma_n}
=(\det A)\epsilon^{\alpha_1\cdots\alpha_n}=\epsilon^{\alpha_1\cdots\alpha_n}.
$$
The lower <tensor> is preserved in the same way using $\det A^{-1}=1$. Thus \b[$\delta$, $\epsilon^{\alpha_1\cdots\alpha_n}$ and $\epsilon_{\beta_1\cdots\beta_n}$ are invariant.]
When $j\geq2$, the permutation of two upper slots commutes with the <tensor product representation>. Its $+1$ and $-1$ <eigenspaces> are nonzero proper <invariant subspaces>. The same argument uses two lower slots when $k\geq2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and <traceless second-rank tensors>, since <tensor contraction> is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the <special unitary group> acts transitively on unit vectors in its fundamental space, so a nonzero <invariant subspace> must be the full space, and the dual has the same property.
A <symmetric tensor> of upper rank $j$ has one component for each multiplicity vector $(r_1,\ldots,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials gives
$$
\boxed{\dim\operatorname{Sym}^jV=\binom{n+j-1}{j}=\frac{(n+j-1)!}{j!(n-1)!}.}
$$
For a <totally antisymmetric tensor>, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. Hence
$$
\boxed{\dim\bigwedge^jV=\binom nj\quad(0\leq j\leq n),\qquad \dim\bigwedge^jV=0\quad(j>n).}
$$
These are the dimensions of the <symmetric power> and <exterior power>, respectively.
For <SU(4)>, the <two-index antisymmetric representation> has complex dimension six. To establish a six-dimensional <real representation>, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize $\epsilon^{1234}=1$ and define the <antilinear map>
$$
(\mathcal JT)^{ab}=\frac12\epsilon^{abcd}\overline{T^{cd}}.
$$
The invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume <tensor> then returns an upper pair. Thus $\mathcal J$ commutes with <SU(4)>. The identity
$$
\epsilon^{abcd}\epsilon_{cdef}=2(\delta^a_e\delta^b_f-\delta^a_f\delta^b_e)
$$
gives $\mathcal J^2=1$. Its fixed space has
$$
T^{12}=\overline{T^{34}},\qquad T^{13}=-\overline{T^{24}},\qquad T^{14}=\overline{T^{23}}.
$$
Three arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The <real structure of the SU4 exterior square> therefore proves
$$
\boxed{\bigwedge^2\mathbb C^4\text{ is the complexification of a real six-dimensional representation}.}
$$
The ordinary Hermitian <norm> restricts to an invariant positive real <inner product> on this fixed space.
<Irreducible representations> classify states because an exact symmetry commutes with the <Hamiltonian>, preserving energy <eigenspaces>. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and <Casimir operators> provide quantum numbers. Approximate <flavour symmetry> gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.
For the light-quark <flavour symmetry> in <Quantum chromodynamics>, the quark flavours form $\mathbf3$. Their <Triple tensor decomposition for the defining sl3 representation> is
$$
\mathbf3^{\otimes3}=\mathbf{10}_{\rm sym}\oplus\mathbf8_{\rm mixed}\oplus\mathbf8_{\rm mixed}\oplus\mathbf1_{\rm antisym}.
$$
This algebraic decomposition alone does not determine the allowed ground-state <baryons>. The <three-quark colour singlet> is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the <Pauli exclusion principle> requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors have
$$
\mathbf2^{\otimes3}=\mathbf4_{\rm sym}\oplus\mathbf2_{\rm mixed}\oplus\mathbf2_{\rm mixed},\qquad \bigwedge^3\mathbb C^2=0.
$$
Here the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its <tensor square> contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.
Equivalently the <symmetric spin-flavour SU6 representation> has
$$
\operatorname{Sym}^3(\mathbb C^3\otimes\mathbb C^2)
=(\mathbf{10},\mathbf4)\oplus(\mathbf8,\mathbf2),\qquad 56=40+16.
$$
Thus the <Pauli constraint on three-quark flavour multiplets> gives the concise spectrum classification
$$
\boxed{\mathbf8\text{ with }J^P=\tfrac12^+,\qquad \mathbf{10}\text{ with }J^P=\tfrac32^+.}
$$
These are the <baryon octet> and <baryon decuplet>. The <three-quark flavour singlet> requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $\mathbf1,\mathbf8,\mathbf{10}$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the <flavour symmetry> approximate rather than exact.
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