= Solution
For the <special unitary group> $SU(N)$, denote a one-column <Young tableau> with $k$ empty boxes by $\mathcal C_k$. Explicitly the columns needed here are
$$
\mathcal C_1=\boxed{\phantom{a}},\quad
\mathcal C_2=\begin{array}{|c|}\hline\phantom{a}\\\hline\phantom{a}\\\hline\end{array},\quad
\mathcal C_3=\begin{array}{|c|}\hline\phantom{a}\\\hline\phantom{a}\\\hline\phantom{a}\\\hline\end{array},\quad
\mathcal C_4=\begin{array}{|c|}\hline\phantom{a}\\\hline\phantom{a}\\\hline\phantom{a}\\\hline\phantom{a}\\\hline\end{array},\quad
\mathcal C_5=\begin{array}{|c|}\hline\phantom{a}\\\hline\phantom{a}\\\hline\phantom{a}\\\hline\phantom{a}\\\hline\phantom{a}\\\hline\end{array}.
$$
A column $\mathcal C_k$ is the <exterior power> $\Lambda^kV$, with <highest weight> $\omega_k$ and dimension $\binom Nk$. Contraction with the invariant volume <tensor> identifies $V^*\cong\Lambda^{N-1}V$ and $\Lambda^2V^*\cong\Lambda^{N-2}V$. This gives all the requested <Dynkin labels>, <Young tableaux>, dimensions and <tensors>:
$$
\boxed{\begin{array}{c|c|c|c|c}
\text{algebra}&R&\text{highest label}&\text{tableau}&\text{dimension and tensor}\\\hline
A_5&\mathbf6&(1,0,0,0,0)&\mathcal C_1&6:\ v^I\\
A_5&\overline{\mathbf6}&(0,0,0,0,1)&\mathcal C_5&6:\ v_I\\
A_5&\mathbf{15}&(0,1,0,0,0)&\mathcal C_2&15:\ T^{IJ}=-T^{JI}\\
A_5&\overline{\mathbf{15}}&(0,0,0,1,0)&\mathcal C_4&15:\ T_{IJ}=-T_{JI}\\\hline
A_4&\mathbf5&(1,0,0,0)&\mathcal C_1&5:\ v^i\\
A_4&\overline{\mathbf5}&(0,0,0,1)&\mathcal C_4&5:\ v_i\\
A_4&\mathbf{10}&(0,1,0,0)&\mathcal C_2&10:\ T^{ij}=-T^{ji}\\
A_4&\overline{\mathbf{10}}&(0,0,1,0)&\mathcal C_3&10:\ T_{ij}=-T_{ji}
\end{array}}
$$
Here $1\leq I,J\leq6$ and $1\leq i,j\leq5$. Lower indices denote the <dual representation>, equivalent to the <conjugate representation> for these <compact groups>. The <SU(6)> $\mathbf{15}$ in this question is an <exterior square>; the <SU(5)> $\mathbf{15}$ in <3/e/solution> is a <symmetric square>. The ambient <group> is essential to distinguish them.
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