Let $G= \{ g=(x,y,t); \quad x,y,t \in \mathbb R\}$ be the Heisenberg group and $H= \{ g=(x,y,t) \in G; \quad x=0\}= \{ h=(0,y,t); \quad y,t \in \mathbb R\}$ be a subgroup of $G$.
I want to know why the one-dimensional representation of $H$ is given by $$ \pi_{\lambda}(h) =\pi_{\lambda}(0,y,t)= e^{i\lambda \, t} $$ for a fixed $\lambda \in \mathbb R^{*}$, and it is not $$ \pi_{\lambda}(h) = e^{i\lambda \, y} \, ?$$ Thank you in advance
You are right, your second character is not equivalent to the previous one. However, there are more one-dimensional unitary representations (characters) of $H$.
Note that $H\simeq \mathbb R^2$ given by $(0,y,t)\mapsto (y,t)$. Hence, $\widehat H\simeq \widehat {\mathbb R^2}\simeq \mathbb R^2$. This last isomorphism is given by $(a,b)\in\mathbb R^2\mapsto \sigma_{(a,b)}$ given by $$ \sigma_{(a,b)}(y,t)= e^{i(ay+bt)}. $$