Unitary groups and infinitesimal transformations - Schwingers way of deriving Lie groups

In Schwinger's source theory book, he suggests if $G_a$ are the hermitian generators of the Unitary group, then we have an infinitesimal transformation is given by :

$$G = \sum_{a=1}^n \delta\lambda_a G_a$$ Now if we subject to the infinitesimal transformation operator to arbitrary unitary transformation of the group ($\{\lambda\}$ are the group parameters) , $$U^{-1}G_aU = u(\lambda)G_b$$ Alternatively, $$UG_aU^{-1} = G_bv(\lambda)$$ And then, these two quantities are related by $v = (u^T)^{-1}$

Firstly, I am not able to see why the author is trying to do this kind of unitary transform on the group element, Is it possibly to get a representation? How further do we proceed with this.

Answer

As I mentioned in my comment, I believe you are talking about the adjoint representation of a Lie Group $G$ with a Lie algebra $\mathfrak{g}$:

$$\begin{equation} \forall x \in \mathfrak{g}, \;\; \mathrm{Ad} \: D(g) : x \mapsto D(g) x D^{-1}(g) \in \mathfrak{g} \end{equation}$$ where $D(g)$ denotes a represention of $g \in G$. One method you can see why $D(g) x D^{-1}(g) \in \mathfrak{g}$ is by considering the above transformation: $$\begin{equation} \begin{aligned} x' & = D(g) x D^{-1}(g) \\ &= e^{i \lambda^i G_i} x e^{-i \lambda^i G_i} \\ &= (1 + i \lambda^i G_i) x (1 - i \lambda^i G_i) + O(\lambda^2) \\ &= x - x i \lambda^i G_i + i \lambda^i G_i x + O(\lambda^2)\\ &= x + i \lambda^i [G_i, x] +O(\lambda^2) \\ &= x + i \lambda^i x^j [G_i,G_j] +O(\lambda^2)\\ &= x - \lambda^i x^j f_{ij}{}^{k} G_k +O(\lambda^2)\\ &= (x^k- \lambda^i x^j f_{ij}{}^{k})G_k +O(\lambda^2)\\ &= x'^i G_i +O(\lambda^2) \end{aligned} \end{equation}$$ where: $$\begin{equation} x'^i \equiv x^i- \lambda^j x^k f_{jk}{}^{i} \end{equation}$$ This shows that $x'$ can also be expressed in terms of the generators. In other words, the above transformation ensures that if $x$ lives in the Lie algebra formed by the generators, then $x'$ also lives in that vector space.

Furthermore, note that from the fifth line of the above equation, we can also write:

$$\begin{equation} x' = x + i \lambda^i [G_i, x] = x+ i \mathrm{ad} \; \lambda(x) = e^{i \mathrm{ad} \; \lambda} x \end{equation}$$ Thus: $$\begin{equation} \mathrm{Ad} \: D(g) (x) = e^{i \mathrm{ad} \; \lambda} x \end{equation}$$ which shows the relation between the adjoint representation of the Lie group and the adjoint representation of the Lie algebra.