= Solution
For the embedding in <4/a/solution>, restriction retains the first four <Dynkin label> coordinates, since the <simple coroots> of $A_4$ are the first four of $A_5$. In a coordinate-free version of the same projection, let $\eta_i$ be the five defining <weights of a representation> of $A_4$. Restriction gives $\varepsilon_i\mapsto\eta_i$ for $1\leq i\leq5$, while $\varepsilon_6\mapsto0$.
For the <fundamental representation>, its first five <weights of a representation> therefore form a $\mathbf5$ and the sixth is the zero <weight of a representation> of a <trivial representation>. Negating the <weights of a representation> gives the <conjugate representation>:
$$
\boxed{\mathbf6\downarrow_{SU(5)}=\mathbf5\oplus\mathbf1,\qquad
\overline{\mathbf6}\downarrow_{SU(5)}=\overline{\mathbf5}\oplus\mathbf1.}
$$
For the <two-index antisymmetric representation>, split the fifteen <weights of a representation> into the ten with $i<j\leq5$ and the five with $j=6$. Their restricted values are $\eta_i+\eta_j$ and $\eta_i$, respectively. These are precisely the <weight systems> of $\mathbf{10}$ and $\mathbf5$. Negating again gives
$$
\boxed{\mathbf{15}\downarrow_{SU(5)}=\mathbf{10}\oplus\mathbf5,\qquad
\overline{\mathbf{15}}\downarrow_{SU(5)}=\overline{\mathbf{10}}\oplus\overline{\mathbf5}.}
$$
For instance the highest $A_5$ <weight of a representation> $(0,1,0,0,0)$ restricts to $(0,1,0,0)$ for the $\mathbf{10}$, while $(1,0,0,0,-1)$ restricts to $(1,0,0,0)$ for the $\mathbf5$. The restricted <weight systems> are disjoint unions of multiplicity-one sets in these cases, and their dimensions check as $6=5+1$ and $15=10+5$.
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