The Heisenberg group $H_1$ is the set $\mathbb C\times \mathbb R$ endowed with the group law
$$ (z,t)\cdot(w,s) =\left (z+w, \,t+s+\tfrac{1}{2}\Im m(z \bar{w})\right); \quad \forall z,w \in \mathbb C\,\text{and}\, \,t,s\in \mathbb R.$$

Let $\delta$ be the Dirac distribution supported at $e=(0,0)$ the identity element on $H_1$. If D is a left-invariant differential operator on $H_1$, we have $Df = D(f *\delta) = f*D\delta$; by the usual
abuse of language we write this as $f * D$. Let $Z$ the left-invariant
vector field given by
$$\begin{equation}
f*Z=\left(\frac{\partial}{\partial z}+i\overline{z}\frac{\partial}{\partial t}\right)f. \qquad (1)
\end{equation}$$

Using the function $A$ on $H_1$ given by $A=|z|^2+it$, we define the map $h$ on $H_1-\left\{e\right\}$ by
$$ h(z,t)=\left(\frac{-z}{|z|^2-it},\frac{-t}{|z|^4+t^2}\right)=\left(-z\overline{A^{-1}},-t A^{-1} \overline{A^{-1}}\right).$$ 

By the use of the following identities
$$\begin{equation}
z*Z=1,\,\,\, \overline{z}*Z=0,\,\,\, t*Z=i\overline{z}. \qquad\qquad\qquad (1.1)\\
A*Z=0,\quad \overline{A}*Z=2\overline{z}.\qquad\qquad\qquad\qquad\qquad   \qquad   (1.2)\\
(z\circ h)*Z=\overline{A^{-2}} \, (2|z|^2-\overline{A}), \quad (t\circ h)*Z=-i\overline{z}\, \overline{A^{-2}}. \quad (1.3)
\end{equation}$$
By a simple calculation we found the equations (1.1) and (1.2), and for (1.3) see the answer of @F Zaldivar below.

Finally, I ask if someone can help me to check the following  
$$\begin{equation}
(f\circ h)*Z=-\overline{A^{-1}}\left|(f*Z)\circ h+2\overline{z}\, (f*E_{z})\circ h\right|; \quad E_{z}:= zZ. \quad (2)
\end{equation}$$

$\triangle$ I found these formulas, in the paper of: Korányi, Kelvin transforms and harmonic polynomials on the Heisenberg group. JFA 1982. 
The result is given in the case of $H_n=\mathbb C^n\times \mathbb R$, and I tried to do it again in the case $H_1=\mathbb C\times \mathbb R$ (n=1).

Thank you in advance