N&=&\left\{N,\;\bar\varepsilon\Omega\right\}
   =\bar\varepsilon\frac{\delta\Omega}{\delta P_N}\nonumber\\
&=&\bar\varepsilon\left[-\frac{\partial{\cal H}}{\partial P_N}\theta^0
   -\frac{\partial f}{\partial V}V'\theta^r
   -\frac{\partial f}{\partial W}W'\theta^r
   -\frac{\partial f}{\partial V} V N^r(\theta^0)'
   -\frac{\partial f}{\partial V} V(\theta^r)'-P_{\bar\theta_0}\right]\nonumber\\
\label{delN2}
&=&\bar\varepsilon\left[-\dot N\theta^0-N'\theta^r-N\dot\theta^0+N N^r(\theta^0)'\right].\end{aligned}$$ Here we used one of the Hamiltonian equations $\dot N=\displaystyle\frac{\delta H}{\delta P_N}$ (\[DN\]), and the expression for $P_{\bar\theta_0}$ (\[Pbt0\]). To check (\[dell0\]), (\[dellr\]) one should firstly find $\delta P_N$, $\delta P_{N^r}$. $$\begin{aligned}
\delta P_N&=&\left\{P_N,\;\bar\varepsilon\Omega\right\}
   =-\bar\varepsilon\frac{\delta\Omega}{\delta N}
   =\bar\varepsilon\left[-\frac{\partial\Omega}{\partial N}
    +\left(\frac{\partial\Omega}{\partial N'}\right)'\right]\nonumber\\
&=&\bar\varepsilon\left[\frac{\partial{\cal H}}{\partial N}\theta^0
    -\left(\frac{\partial{\cal H}}{\partial N'}\theta^0\right)'
    +\frac{W W'(\theta^0)'}V\right]\nonumber\\
&=&\bar\varepsilon\left[\left(\frac{\partial{\cal H}}{\partial N}
    -\left(\frac{\partial{\cal H}}{\partial N'}\right)'\right)\theta^0
    -\frac{\partial{\cal H}}{\partial N'}(\theta^0)'
    +\frac{W W'(\theta^0)'}V\right]\nonumber\\
&=&\bar\varepsilon\left[-\dot P_N\theta^0-\left(\frac{W W'}V
    -\frac1N\bar P_{\theta^0}\theta^r
    +\frac{N^r}N\bar P_{\theta^r}\theta^r\right)(\theta^0)'
    +\frac{W W'(\theta^0)'}V\right]\nonumber\\
\label{delPN}
&=&\bar\varepsilon\left[-\dot P_N\theta^0
    +\frac1N\left(N\dot{\bar\theta_0}+N^r\dot{\bar\theta_r}\right)\theta^r(\theta^0)'
    -\frac{N^r}N\dot{\bar\theta_r}\theta^r(\theta^0)'\right]
    =\bar\varepsilon\left[-\dot P_N\theta^0+\dot{\bar\theta_0}\theta^r(\theta^0)'\right];\\
\label{delPNr}
\delta P_{N^r}&=&\bar\varepsilon\left[-\dot P_{N^r}\theta^0+\dot{\bar\theta_r}\theta^r(\theta^0)'\right].\end{aligned}$$ Here we also used the Hamiltonian equation $\dot P_N=-\displaystyle\frac{\delta H}{\delta N}
=-\displaystyle\frac{\partial{\cal H}}{\partial N}+\left(\displaystyle\frac{\partial{\cal H}}{\partial N'}\right)'$ (\[DPN\]). Keeping in mind the relations between $P_N$, $P_{N^r}$ and $\lambda_0$, $\lambda_r$ (\[lambda\]), one can be convinced that Eqs. (\[dell0\]), (\[dellr\]) are correct.

One can find transformations for $\delta P_V$: $$\begin{aligned}
\delta P_V&=&\bar\varepsilon\left[-\dot P_V\theta^0-P'_V\theta^r
   -P'_N\frac{\partial f}{\partial V}\theta^r
   -P'_{N^r}\frac{\partial f^r}{\partial V}\theta^r
   +P_N\frac{\partial^2 f}{\partial V^2}V(\theta^r)'
   +P_{N^r}\frac{\partial^2 f^r}{\partial V^2}V(\theta^r)'\right.\nonumber\\
&+&P_N\frac{\partial^2 f}{\partial V^2}V N^r(\theta^0)'
   +P_{N^r}\frac{\partial^2 f^r}{\partial V^2}V N^r(\theta^0)'
   +\frac{\partial f^r}{\partial V}\bar P_{\theta^r}(\theta^0)'\theta^r\nonumber\\
\label{delPV}
&+&\left.\frac1N\frac{\partial f}{\partial V}\bar P_{\theta^0}(\theta^0)'\theta^r
   -\frac{N^r}N\frac{\partial f}{\partial V}\bar P_{\theta^r}(\theta^0)'\theta^r
   -\frac{N W W'}{V^2}(\theta^0)'\right].\end{aligned}$$ The transformation of (\[PV\]) gives the same result. Similarly one 