Gauge covariant derivative of an adjoint action: $\psi(x) \to g \psi(x) g^{-1}$, instead of a left action $\psi(x)\to e^{iq\theta(x)} \psi(x)$

In the case where the transformation on $$\psi$$ is applied from the left:

$$\psi(x)\to e^{-iq\theta(x)}\psi(x).$$

The gauge covariant derivative is

$$D_\mu = \partial_\mu - iqA_\mu \tag{1}$$

and the field is given as follows:

$$F_{\mu\nu}=[D_\mu,D_\nu]. \tag{2}$$

My question is; what are the equivalents to equation (1) and (2) if we have an adjoint action such as this

$$\psi(x) \to g(x)\psi(x)g^{-1}(x)$$

where $$g(x)$$ could be arbitrary general linear transformations for instance. Does the use of a adjoint action transformation significantly changes (1) and (2)?

First, let me clear a possible confusion: as you've tagged the question as "electromagnetism", I assume you're going for an $$U(1)$$ symmetry. In this case, the adjoint does not transform, as the group is abelian (you can see it because if $$\psi(x)$$ is in an irreducible representation, then it's just a complex function, not a vector: $$g$$ and $$\psi$$ commute, so the transformation is the identity). So your question is trivial for electromagnetism: the covariant derivative of a field $$\psi(x)$$ in the adjoint representation is just the standard derivative.
But let's see the answer for a non abelian group anyway. A field $$\psi(x)$$ in the adjoint representation will be a linear combination of the generators $$T^a$$ of the adjoint representation (given by the structure constants). How do you get the covariant derivative of such a thing? Well, from the transformation itself!
What's the general transformation for a field in any representation, spanned by the generators $$T^a$$ of the adjoint representation? A general field in the representation can be written in components as $$\psi(x)=\psi^a(x) T^a.$$ Component fields $$\psi^a$$ have by definition the following derivative: $$(D_\mu\psi)^a(x)=\partial_\mu\psi^a(x)-\mathrm i g (A_\mu^c(x)T^c)^{ab}\psi^b(x).$$ The term on the right is basically a multiplication of the connection matrix $$A_\mu^c(x)T^c$$. I'm here using the fact that the adjoint representation has the same dimension as the dimension of the Lie algebra.I will keep all indexes up but for the spacetime index $$\mu$$.
Now, contract what you got with the generators, and let's keep all indexes out: you get $$(D_\mu\psi)^a(T^a)^{bc}=(\partial_\mu \psi^a)(T^a)^{bc}-\mathrm i g A_\mu^d(T^d)^{ae}(T^a)^{bc}\psi^e.$$ Now, we use the fact that the generators are the structure constants, $$(T^a)^{bc}=-\mathrm i f^{abc}$$. You can rewrite the product of $$T$$'s as $$(T^d)^{ae}(T^a)^{bc}=-f^{dae}f^{abc}=f^{dac}f^{eba}-f^{eac}f^{dba}=(T^e)^{ac}(T^d)^{ba}-(T^d)^{ac}(T^e)^{ba}=[T^d,T^e]^{cb}.$$ Here I've used antisymmetry and Jacobi identity on the structure constants. Please check my signs!
Let's return to our expansion. $$(D_\mu\psi)^a(T^a)^{bc}=\partial_\mu\psi^a(T^a)^{bc}-\mathrm i g A_\mu^d[T^d,T^e]^{bc}\psi^e.$$ In matrix representation, we get $$D_\mu\psi=\partial_\mu\psi-\mathrm i g[A_\mu,\psi].$$ This is the generalization of (1) to fields in the adjoint representation. The definition of the field strength tensor remains unvaried.