= Solution
For $SU(4)$, <Dynkin labels> $(a,b,c)$ correspond to a <Young diagram> with row lengths $(a+b+c,b+c,c)$. Numbered boxes below give representative <Young tableaux>; they specify the shapes, not additional component values:
$$
\begin{array}{c|c|c|c}
\text{Dynkin labels}&\text{tableau}&\text{dimension}&\text{tensor realization}\\\hline
(1,0,0)&\boxed{1}&4&v^i\\[5pt]
(0,1,0)&\begin{array}{l}\boxed{1}\\[-3pt]\boxed{2}\end{array}&6&A^{ij}=-A^{ji}\\[8pt]
(0,0,1)&\begin{array}{l}\boxed{1}\\[-3pt]\boxed{2}\\[-3pt]\boxed{3}\end{array}&4&B^{ijk}=B^{[ijk]}\ \leftrightarrow\ b_i\\[11pt]
(1,0,1)&\begin{array}{l}\boxed{1}\,\boxed{2}\\[-3pt]\boxed{3}\\[-3pt]\boxed{4}\end{array}&15&X^i{}_j,\quad X^i{}_i=0
\end{array}
$$
Here the <dimensions> are $4$, $\binom42=6$, $\binom43=4$, and $4^2-1=15$. The third is the <dual representation> $\overline{\mathbf4}$: the invariant volume <tensor> identifies $b_l=\epsilon_{lijk}B^{ijk}/3!$. The last is the <Adjoint representation>, equivalent to the shape $(2,1,1)$ <Young symmetrizer> construction after converting its three-box column to one lower index and removing the <trace>.
For $U\in SU(4)$, the <tensor> transformation laws are
$$
\boxed{\begin{aligned}
v'^i&=U^i{}_jv^j,\\
A'^{ij}&=U^i{}_kU^j{}_lA^{kl},\\
B'^{ijk}&=U^i{}_lU^j{}_mU^k{}_nB^{lmn},\\
b'_i&=b_j(U^{-1})^j{}_i,\\
X'^i{}_j&=U^i{}_kX^k{}_l(U^{-1})^l{}_j.
\end{aligned}}
$$
These preserve antisymmetry and tracelessness. The <Adjoint representation> can be treated as complex traceless <matrices>; a real adjoint <scalar field> is conventionally represented by Hermitian traceless <matrices> with the same conjugation action.
Adding one box to the one-box <Young diagram> gives row shapes $(2)$ and $(1,1)$. Thus the first <tensor product> splits into the <symmetric square> and <exterior square>:
$$
\boxed{(1,0,0)\otimes(1,0,0)=(2,0,0)\oplus(0,1,0),\qquad
\mathbf4\otimes\mathbf4=\mathbf{10}\oplus\mathbf6.}
$$
Their <tensors> are respectively $S^{ij}=S^{ji}$ and $A^{ij}=-A^{ji}$. Adding one box to the three-box column gives shapes $(2,1,1)$ and $(1,1,1,1)$. The full four-box column is the <determinant representation>, trivial because $\det U=1$. Consequently
$$
\boxed{(1,0,0)\otimes(0,0,1)=(1,0,1)\oplus(0,0,0),\qquad
\mathbf4\otimes\overline{\mathbf4}=\mathbf{15}\oplus\mathbf1.}
$$
For a mixed <tensor> $T^i{}_j$, the corresponding pieces are $T^i{}_j-\tfrac14\delta^i{}_jT^k{}_k$ and $\tfrac14\delta^i{}_jT^k{}_k$.
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