I guess by "bosons" you're referring to gauge bosons?
If so then start with some matter field $ \psi(x)$ which transforms under the gauge group. For local gauge transformations the gauge group element $g$ is spacetime dependent $g(x)$, and the transformation is
$$\psi(x) \longrightarrow \psi'(x) = g(x)\psi(x).$$
Derivatives would transform as
$$\partial_{\mu}\psi(x) \longrightarrow g(x)\partial_{\mu}\psi(x)+(\partial_{\mu}g(x))\psi(x),$$
i.e. inhomogeneously. We would like a gauge covariant derivative $D_{\mu}$ which transforms homogeneously as
$$D_{\mu}\psi(x) \longrightarrow g(x)D_{\mu}\psi(x).$$
To achieve this, we define
$$D_{\mu}\psi = \partial_{\mu}\psi - A_{\mu}\psi,$$
where $A_{\mu} = \mathbf{A}_{\mu} \cdot\boldsymbol{{\tau}}$ and $\boldsymbol{\tau}$ are the generators of the Lie algebra of the gauge group and $A_{\mu}$ is our bosonic gauge field. This introduction of gauge bosons via the derivative term is sometimes referred to as minimal coupling.
In order to achieve this, $A_{\mu}$ is forced to have the transformation law
$$A_{\mu} \longrightarrow A'_{\mu} = gA_{\mu}g^{-1} + (\partial_{\mu}g)g^{-1}.$$
Just looking at how the $A_{\mu}$ are transforming under the group action (the first term), we recognize the adjoint representation.
Of course, on the global stage, the fields $\psi$ can be interpreted as bundle sections and the gauge fields as bundle connections. $A_\mu$'s transformation law will be recognisable as a transformation of connection coefficients under the action of the bundle's structure group. A good reference is Nakahara, or this link.
Best 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.