Suppose that $A$ is a invertible matrix with complex entries such that $\{\|A^n\|:n\in\mathbb{Z}\}$ is bounded. We consider the Jordan normal form of $A$. It is well known that \begin{align*} &\begin{pmatrix} x & 1 & 0 &\cdots & 0 & 0\\ 0 & x & 1 &\cdots & 0 & 0\\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots\\ 0 & 0 & 0 &\cdots & x & 1\\ 0 & 0 & 0 &\cdots & 0 & x\\ \end{pmatrix}^n_{d\times d} = \begin{pmatrix} x^n & (x^n)' & (x^n)'' &\cdots & (x^n)^{(d-2)} & (x^n)^{(d-1)}\\ 0 & x^n & (x^n)' &\cdots & (x^n)^{(d-3)} & (x^n)^{(d-2)}\\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots\\ 0 & 0 & 0 &\cdots & x^n & (x^n)'\\ 0 & 0 & 0 &\cdots & 0 & x^n\\ \end{pmatrix}\\ =&\begin{pmatrix} x^n & nx^{n-1} & n(n-1)x^{n-2} &\cdots & n\cdots(n-(d-2)+1)x^{n-(d-2)} & n\cdots(n-(d-1)+1)x^{n-(d-1)}\\ 0 & x^n & nx^{n-1} &\cdots & n\cdots(n-(d-3)+1)x^{n-(d-3)} & n\cdots(n-(d-2)+1)x^{n-(d-2)}\\ \vdots & \vdots & \vdots &\ddots & \vdots & \vdots\\ 0 & 0 & 0 &\cdots & x^n & nx^{n-1}\\ 0 & 0 & 0 &\cdots & 0 & x^n\\ \end{pmatrix},\quad n\in\mathbb{Z}, \end{align*} so the $n$-th power of a size-$d$ Jordan block with digonal element $x$ has $\ell^1$ or $\ell^\infty$ norm $$ |x|^n+n|x|^{n-1}+\cdots+n\cdots(n-(d-1)+1)|x|^{n-(d-1)}, $$ and this number is bounded with respect to $n$ if and only if $|x|=1$ and $d=1$, so the Jordan form of $A$ is a unitary matrix.
What happens in infinite dimension? Let $V$ be a Hilbert space and $T\in\operatorname{GL}(V)$ (in the sense that $T$ is invertible and $T$, $T^{-1}$ are continuous). If $\{\|T^n\|:n\in\mathbb{Z}\}$ is bounded, must $T$ be conjugated to an orthogonal operator? Also, if $V$ is merely a Banach space, could there be anything interesting to be said?
If $T$ is an invertible operator on Hilbert space such that $\sup \{\|T^n\|: n \in \mathbb{Z}\}$ finite (the usual terminology for this is might be that $T$ is "power-bounded" - or maybe this terminology only captures the $n \geq 0$ part of your hypothesis, I'm not sure), then there is an invertible $Q$ such that $QTQ^{-1}$ is unitary. Indeed, $Q$ can be taken to be self-adjoint and satisfying $k^{-1} \mathbf{1} \leq Q \leq k \mathbf{1}$, where $k = \sup \{\|T^n\|: n \in \mathbb{Z}\}$.
In this form, the result is due to Béla Szőkefalvi-Nagy. See Sz.-Nagy, On uniformly bounded linear transformations in Hilbert space, Acta Sci. Math. (Szeged) 11 (1947), 152-157. The original paper is freely available online and is very readable. The idea is to take a Banach limit $L$ on the space of bounded numerical sequences, and to define a sesquilinear form $[\cdot, \cdot]$ on the Hilbert space by way of $[\xi, \eta] := L(n \mapsto \langle T^n \xi, T^n \eta \rangle)$, where $\langle \cdot, \cdot \rangle$ is the inner product on the Hilbert space. It is not hard to see that there is a self-adjoint operator $A$ satisfying $[\xi,\eta] = \langle A \xi, \eta \rangle$ for all $\xi$ and $\eta$, and that then $Q := A^{1/2}$ will do the job.
It may be that more modern treatments have shed more light on this and generalizations; I seem to recall Vern Paulsen's Completely Bounded Maps and Operator Algebras covering some of this type of thing from a more modern point of view (or at least, a point of view that post-dates 1947). But I do not have the book with me. In particular, I do not know what happens if $V$ is only a Banach space.