Here's a classical construction.

Fix a sequence $(s_n)_{n\in\mathbf{N}}$ of nonzero real numbers in $[-\pi,\pi]$. On $\mathbf{C}^\mathbf{N}$, define a linear automorphism by $T((x_n))=(e^{is_n}x_n)$. Note that the unique fixed point of $T$ is 0. Endow $\mathbf{C}^\mathbf{N}$ with the $\ell^2$-"norm", allowing infinite values. Then $T$ preserves this norm. In particular, it permutes the cosets of $\ell^2$, the subspace of elements in $\mathbf{C}^\mathbf{N}$ with finite norm, and the restriction of $T$ to any of this coset is an isometry.

Now assume that $(s_n)$ is in $\ell^2$. Since $e^{is_n}=1+is_n+o(s_n)$, we deduce that $T(1)-1\in\ell^2$. Thus $T$ maps the coset $V=1+\ell^2$ into itself, and thus induces a (surjective) self-isometry of $V$.

Define a length function on $\mathbf{Z}$ by $f(n)=\|T^n(1)-1\|$. (Length function is the axioms that you call norm, except that nonzero elements are allowed to have zero length). Since no power of $T$ fixes 1, we see that this length is indeed nonzero outside zero.

Since $T$ has no fixed point and $V$ is isometric to a Hilbert space, $T$ has unbounded orbits (by Jung's center lemma); in particular, the above length function is unbounded.

To ensure that the length is non-discrete, we need stronger hypotheses on $(s_n)$. Say, $s_n=2\pi/(\max(2,n)!$. Then it is easy to check that $T^{n!}(1)$ tends to 1, and hence the length $f(n!)$ tends to 0.

If one wishes a more explicit argument argument avoiding the center lemma, one can replace the constant sequence 1 by the sequence c=(n), and keep the same choice of $(s_n)$, so $\|T(c)-c\|<\infty$, and define $f'(n)=\|T^nc-c\|$. Then the $n$-th coordinate of $T^{n!/2}c$ is $-n$. So for $n\ge 2$, $f'(n!/2)=\|T^{n!/2}c-c\|\ge 2n$, and hence $f'$ is unbounded. The argument showing that the length of $n!$ tends to zero is essentially the same computation.

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