Yes, it is "standard" that a finite-dimensional vector space over a complete, non-discrete, division algebra (!) has a unique topology compatible with a topological vector space structure (=Hausdorff, addition is continuous, scalar multiplication is continuous). This is probably proven (at least for real or complex scalars, but the proofs should generalize) in any source on general TVS's, e.g., section 5, my notes. Without completeness, or without discreteness, there are easy counter-examples.
For sake of completeness, I am writing a full answer based on the suggestion of
Pietro Majer.
The following are equivalent:
1) $A$ is an isometry w.r.to some norm.
2) $A$ is diagonalizable (over $\mathbb{C}$) , with all eigenvalues of modulus 1.
3) All orbits of $A$ are bounded ( $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$ for any $x\in \mathbb{R}^n$).
4) $A$ is an isometry w.r.to some inner product.
$(1)\iff (4)$ leads to an interesting point: The union of all isometries of all norms equals the union of all isometries of all inner products.
Proof:
$(1) \Rightarrow (2):$
Assume $||$ is a norm preserved by $A$. Then the operator norm of $A$ w.r.t to $||$ equals 1. Also $||A^n||_{op}=1$. By the spectral radius formula: $\rho(A)=\lim_{n\rightarrow\infty}||A^n||^{1/n}=1 $. The same argument for $A^{-1}$ implies $\rho(A^{-1})=1$. This implies all the eigenvalues (including the complex ones) are of absolute value one.
Also, it is easy to see that If $A$ is an isometry of the norm $| |_1$ , $P∈GL(\mathbb{R^n})$, $P^{−1}AP$ is an isometry of the norm $||_2$ where $||x||_2=||Px||_1$. Thus, the property that a given matrix admit such a norm is invariant under similarity.
So it is enough to focus upon matrices of Jordan form. (which is available to us since we work over $\mathbb{C}$).
Now assume $A$ is not diagonalizable. By looking at Jordan form of a non-diagonalizable matrix, we can see there is a vector $v∈\mathbb{C}^n$ such that $||A^kv||_{Euclidean}$ diverges. (Look at the example of $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}$ given in the question).
Since all the norms are equivalent This implies that $||A^kv||$ diverges. But since $v=x+iy$ with $x$ and $y$ in $\mathbb{R}^n$, and $A^kv=A^kx+iA^ky$, either $||A^kx||$ or $||A^ky||$ diverge. This is impossible if $A$ is an isometry of $||$.
$(2) \Rightarrow (4):
$ This is proved here (The basic idea is to look at each Jordan block separately).
$(4) \Rightarrow (1):$ Obvious.
It remains to prove $(1) \iff (3)$:
$(3) \Rightarrow (1):$ If all orbits of $A$ are bounded, then we can renorm $\mathbb{R}^n$ by $\|x\|_A:=\sup_{k\in \mathbb{Z}} \|A^k x\|$, which is an $A$-invariant equivalent norm.
$(1) \Rightarrow (3):$ This follows immediately by the fact all norms on a finite dimensional vector space are equivalent. The orbits of $A$ are all of constant norm ($\|x\|$) w.r.t to the norm $A$ preserves, hence are bounded. (w.r.t any other norm).
Best Answer
Field $\mathbb Q$ with the usual absolute value $|\cdot|$ from the real numbers.
Two norms on $\mathbb Q^2$ ... $$ \|(x,y)\|_1 = |x|+|y| $$ and $$ \|(x,y)\|_2 = \left|\,x+\sqrt{2}\;y\,\right| $$ In both of these, $|\cdot|$ is still the usual absolute value for the real numbers.