Check the stability of the system, $y[n]=u[n]$
(Long Comment) Sorry guys, I did not get why that system created so many conceptual problems to you. For instance, consider the unforced system $y(n)=0$. If I define $\pi:\{0\}\times\mathbb Z\to\{0\}$, $\pi(y,n)=0$ then $(\{0\},\mathbb Z,\pi)$ defines a dynamical system on the state space $\{0\}$. In fact $\pi$ is continuous, and it satisfies the identity axiom ($\pi(x,0)=x$ for all $x\in\{0\}$) and the group axiom ($\pi(\pi(x,n_1),n_2) = \pi(x,n_1+n_2)$ for all $(x,n_1),(x,n_2)\in\{0\}\times\mathbb Z$).

I agree that this is a degenerate case. I guess I can extend the state space to $\mathbb R$ by defining the system in a different way. For instance, let 
$$
x(n+1)=0\qquad (1)
$$ 
be a ordinary difference equation and define $\pi$ as the flow of (1). Then $(\mathbb R,\mathbb N,\pi)$ defines a dynamical system on $\mathbb R$. Take the forced system
\begin{align*}
x(n+1) &amp;= 0\cdot x(n) + 0\cdot u(n)\\
y(n) &amp;= 0\cdot x(n) + 1\cdot u(n)\qquad\qquad (2)
\end{align*}
That makes sense to me. It is a LTI system with zero relative degree. To check BIBO stability let take a reference input $\bar u:\mathbb N\to\mathbb R$ and a perturbation $\delta u:\mathbb N\to\mathbb R$. Let $u(n)=\bar u(n)+\delta u(n)$.
Let $\bar y:\mathbb N\to\mathbb R$ be the solution corresponding to the input $\bar u$ and let $y:\mathbb N\to\mathbb R$ be the one corresponding to the input $u$. Then
\begin{align*}
\|y(n)-\bar y(n)\| = \|u(n)-\bar u(n)\|=\|\delta u(n)\|
\end{align*}
So trivially, for any $\epsilon&gt;0$ there exists $\delta_\epsilon&gt;0$ such that
$\|\delta u(n)\|\le \delta_\epsilon\implies \|y(n)-\bar y(n)\|\le\epsilon$. You can see this also as a continuity property of the solutions. In fact, you can define the space $Y$ of all the bounded output trajectories $y:\mathbb N\to\mathbb R$ and the space $U$ of all the bounded input ones. Let endow $Y$ with the norm $\|y\|:=\sup_{n\in\mathbb N}|y(n)|$ and $U$ with the norm $\|u\|:=\sup_{n\in\mathbb N}|u(n)|$. Then the system (2) definest an operator $T:U\to Y$, and asking for BIBO stability is the same as asking for the continuity of $T$ in the sup norms.