 Solution

Use the standard chain ordering of the five simple roots of $SU(6)$. A Young diagram of row lengths $r_1\ge \cdots \ge r_6=0$ has highest weight Dynkin labels $a_i=r_i-r_{i+1}$. A full column of six boxes can be removed because it is the trivial determinant representation. The required data areIn the same order, the Young tableaux areEach upper index transforms by $U$, and each lower index by $U^{-1}$. In particular $M\mapsto UM{U}^{-1}$ is the complex Adjoint representation. The column of five boxes can be represented either by $w_i$ or by the totally antisymmetric tensor $B^{i_1\dots i_5}$, using the invariant six-index alternating tensor:Thus it really is the conjugate representation, rather than a second independent six-dimensional choice. Similarly converting the lower index of $M^i{}_j$ into five alternating upper indices gives the displayed hook Young diagram; the removed trace would be its full six-box column component.

The dimension counts areThe first two count independent entries of symmetric tensors and antisymmetric tensors; the last removes one trace from a mixed tensor. The hook-content formula independently gives these dimensions from the Young diagrams.