Its commonly said that gauge bosons transform according to the adjoint representation of the corresponding gauge group. For example, for SU(2) the gauge bosons live in the adjoint 3 dimensional representation and the gluons in the 8 dimensional adjoint of SU(3).
Nevertheless, they transform according to
A_μ→A′_μ=UA_μU^\dagger−ig(∂_μU)U^\dagger ,
which is not the transformation law for some object in the adjoint representation. For example the W bosons transform according to
(W_μ)_i=(W_μ)_i+∂_μa_i(x)+\epsilon_{ijk}a_j(x)(W_μ)_k.
Answer
A gauge field transforms in the adjoint of the gauge group, but not in the adjoint (or any other) representation of the group of gauge transformations.
In detail:
Let G be the gauge group, and \mathcal{G} = \{g : \mathcal{M} \to G \vert g \text{ smooth}\} the group of all gauge transformations.
A gauge field A is a connection form on a G-principal bundle over the spacetime \mathcal{M}, which transforms as A \mapsto g^{-1}Ag + g^{-1}\mathrm{d}g for any smooth g : \mathcal{M} \to G. If g is constant, i.e. not only an element of \mathcal{G}, but of G itself, this obviously reduces to the adjoint action, so A does transform in the adjoint of G, but not in the adjoint of \mathcal{G}. With respect to \mathcal{G}, it does not transform in any proper linear (or projective) representation in the usual sense, but like an element of a Jet bundle.
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