The action should be invariant under the action of gauge group, therefore, you have to construct something, that when considering the action of gauge group remains unchanged.
Assume that the field stregth tensor $F$ transform under some representation of $G$:
$$
F \Rightarrow U^{i_1 \ldots i_n}_{j_1 \ldots j_m} F
$$
To make this more clear, let us list several simple examples:
$$
\phi^i \Rightarrow U_{j}^{i} \phi^j - \text{fundamental representation}
$$
$$
\phi_i \Rightarrow U_{i}^{j} \phi_j - \text{antifundamental representation}
$$
$$
\phi_i^{j} \Rightarrow U_{i}^{a} \phi_a^{b} U_{b}^{j} - \text{adjoint representation}
$$
After multiplying $F$ and $F^{*}$ and taking of the trace, there has to remain and object without free indices. The operation of multiplication and $\text{Tr}$ contracts equal number of lower and upper indices, so the only consistent way to make singlet is to have $m = n$.
$$
\text{Tr} \ F \wedge F^{*} = \text{Tr} \ U^{i_1 \ldots i_m}_{j_1 \ldots j_m} U^{j_1 \ldots j_m}_{k_1 \ldots k_m} \ F \wedge F^{*} = \text{Tr} \ U^{i_1 \ldots i_m}_{j_1 \ldots j_m} U^{j_1 \ldots j_m}_{i_1 \ldots i_m} \ F \wedge F^{*}
$$
The written by far expression by far doesn't correspond to irreducible representation in general - i.e, one can symmetrize (antisymmetrize) over $i_a$ or $i_j$. I may be wrong, but it seems that the general action of the aforementioned form is some irreducible representation of tensor product of several adjoint representations.
