Let $\mathbb E:=(\mathbb R^4, \cdot)$ be a Carnot group whose Lie algebra is given by $\mathfrak g=V_1\oplus V_2 \oplus V_3$, where $V_1=span\{X_1,X_2\},$ $V_2=span\{X_3\},$ $V_3=span\{X_4\}$, the only non trivial commutation relation beibg: $$[X_1,X_2]=-X_3\quad [X_1,X_3]=-X_4.$$
Now, we know that we can endow $\mathbb E$ with a family of automorphisms $\delta_\lambda$ defined as $$(x_1,x_2,x_3,x_4)\mapsto^{\delta_\lambda}(\lambda x_1, \lambda x_2, \lambda^2 x_3, \lambda^3 x_4), \quad \lambda\in \mathbb R.$$
In exponential coordinates, we know that $X_2=\partial_2-\frac{x_1}{2}\partial_3+\frac{x_1^2}{12}\partial_4.$ Moreover, we denote with $exp:\mathfrak g\ \to \mathbb E$ the exponential map.
We know that $exp(\mathbb RX_2):=\{exp(tX_2):t\in \mathbb R\}$ is a subgroup of $\mathbb E$. I was wondering if $exp(\mathbb RX_2)$ is also a $\delta_\lambda$-homogeneous subgroup, that is $\delta_\lambda(exp(tX_2))\in exp(\mathbb RX_2)$ for each $t\in \mathbb R$.
Is it true? If so, how can I prove it?