This is not a homework exercise. I graduated from univerisity more than 10 years ago. I ask questions from my self-study.
There're two types of symmetry transformations in classical mechanics. One is known as rigid transformations, the other known as gauge transformations.
For the first type, consider a generic action
\begin{equation} S[p,q]=\int dt\left\{p_{i}\dot{q}^{i}-H(p,q)\right\}. \tag{1} \end{equation}
Suppose $Q$ is a conserved charge of (1), then, one has the following infinitesimal symmetry transformations
\begin{align} &\delta_{\epsilon}q^{i}=\epsilon\left\{q^{i},Q\right\}_{PB}=\epsilon\frac{\partial Q}{\partial p_{i}}, \tag{2.a}\\ &\delta_{\epsilon}p_{i}=\epsilon\left\{p_{i},Q\right\}_{PB}=-\epsilon\frac{\partial Q}{\partial q^{i}}, \tag{2.b} \end{align}
where $\epsilon$ is an infinitesimal constant.
For type two, consider a gauge theory
\begin{equation} S[p,q,\lambda]=\int dt\left\{p_{i}\dot{q}^{i}-H_{0}(p,q)-A^{I}\,C_{I}(p,q)\right\}, \tag{3} \end{equation}
where $A^{I}$ are Lagrange multipliers, and $C_{I}$ are first class constraints that generate the following gauge transformations
\begin{align} \delta_{\epsilon}q^{i}&=\epsilon^{I}(t)\left\{q^{i},C_{I}\right\}_{PB}=\epsilon^{I}(t)\frac{\partial C_{I}}{\partial p_{i}}, \tag{4.a} \\ \delta_{\epsilon}p_{i}&=\epsilon^{I}(t)\left\{p_{i},C_{I}\right\}_{PB}=-\epsilon^{I}(t)\frac{\partial C_{I}}{\partial q^{i}}. \tag{4.b} \end{align}
Being first class constraints requires $C_{I}$ to satisfy the following Poisson bracket identities
$$\left\{C_{I},C_{J}\right\}_{PB}=f_{IJ}^{\,\,\,\,K}\,C_{K},\quad\quad\left\{H_{0},C_{I}\right\}_{PB}=h_{I}^{\,\,\,J}\,C_{J},$$
where $f_{IJ}^{\,\,\,\,K}$ is the structure constant of the gauge group. One can check that action (3) is invariant up to a boundary term iff the Lagrange multiplier transforms in the following way
\begin{equation} \delta_{\epsilon}A^{I}=\dot{\epsilon}^{I}(t)-\epsilon^{J}(t)h_{J}^{\,\,\,I}+A^{J}\epsilon^{K}(t)f_{JK}^{\,\,\,\,I}. \end{equation}
Let me denote a finite gauge transformation by $U(t)$, under which $A^{I}$ is transformed to $A^{I}[U]$, and denote the basis of the Lie algebra of the gauge group by $\left\{T_{1},T_{2},\cdots,T_{N}\right\}$. I know that the finite version of the above equation is
$$A^{I}[U]T_{I}:=\mathbb{A}[U]=U^{-1}\dot{U}-U\pmb{h}+U^{-1}\mathbb{A}U.$$
Question 1: What are the finite versions of (2.a), (2.b), and (4.a), (4.b)?
My intuitive guess would be that for any function $f(p,q)$ on the canonical phase space, a finite symmetry transformation would be
$$f(p,q)\rightarrow f(p,q)+\epsilon\left\{f,Q\right\}_{PB}+\frac{1}{2}\epsilon^{2}\left\{\left\{f,Q\right\}_{PB},Q\right\}_{PB}+\mathcal{o}(\epsilon^{3})$$
for the rigid case, and similarly
$$f(p,q)\rightarrow f(p,q)+\epsilon^{I}\left\{f,C_{I}\right\}_{PB}+\frac{1}{2}\epsilon^{I}\epsilon^{J}\left\{\left\{f,C_{I}\right\}_{PB},C_{J}\right\}_{PB}+\mathcal{o}(\epsilon^{3})$$
Question 2: Am I correct? If I'm correct, how to prove it?
