Yang-Mills theory is based on the gauge group $G$ which we take to be $SU(N)$. Consider an example; $$\mathcal{L}=-\frac{1}{4}F^a_{\mu\nu}F^{a\mu\nu}-\sum_{j=1}^N\bar{\psi}_j(i\gamma^\mu D_\mu-m)\psi_j$$ where $\psi_j$ is a Dirac spinor in the fundamental representation.
From here I'll focus on algebras, so $\mathfrak{su}(N)$. A representation of this is a Lie algebra homomorphism $\rho:\mathfrak{su}(N)\rightarrow \mathfrak{gl}(V)$ and $V$ is some vector space.
I understand how things work in the fundamental representation but I'm having trouble seeing how it works when the matter fields are also in the adjoint representation. In this case the fields are now represented by $N\times N$ matricies as well, and there are $N^2-1$ generators.
So for example, if I have a scalar field multiplet $\Phi$ transforming in the adjoint representation I have $$\Phi=\phi^a T^a$$ where $T^a$ are the generators. So I have $N^2-1$ scalar fields $\phi^a$.
But in the case of $\mathcal{N}=4$ $SU(N)$ Super Yang Mills I have six scalars. So it's easy to see a six-component field multiplet st the Lagrangian has $SO(6)$ symmetry, which it does. But all fields still transform in the adjoint rep of $SU(N)$. I do not see how for example this six component multiplet can transform under the adjoint rep. Unless of course, each component of this multiplet is now a field transforming in the adjoint rep.
In the latter case, I do not understand how to expand the fields. Example, take one of my six scalars, $\phi_i(x)$ and expand $\phi_i^a(x)T^a$. Now what are the $\phi_i^a(x)$? The guess would be scalars, but that means my theory actually has $6N^2-6$ scalars instead of 6. Maybe when we say ${\cal N}=4$ SYM has 6 scalars we mean 6 scalar field multiplets transforming the the adjoint rep?
Long story short; How are matter fields represented in the adjoint rep?