Assume $V$ is a finite-dimensional vector space over $\mathbb{R}$, and $T: V \to V$ is a (linear) isomorphism.

When is it possible to construct a norm on $V$ making $T$ an isometry?

(Hopefully, I am looking for necessary & sufficient conditions $T$ should satisfy, i.e. a full characterization of the situation).

**What I have so far:**

A necessary condition: all the real eigenvalues of $T$ are of absolute value $1$. (Since $ T(v)=\lambda v \Rightarrow ||v||=||T(v)||=||\lambda v||=|\lambda| ||v||$ and an eigenvector $v$ must be nonzero).

**This condition is certainly not sufficient:**

For example look at $A$ = $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}: \mathbb{R}^2 \to \mathbb{R}^2$. It is an automorphism which has only one eigenvalue ($\lambda = 1$). However, $A\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+y \\ y \end{pmatrix}$, hence $A^n\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ny \\ y \end{pmatrix}$. Now assume there exist a norm $||$ on $\mathbb{R}^2$ making $A$ an isometry.

In particular $A$ must map the open unit ball $B$ to itself. By properties of norms $B$ must be a bounded open set containing the origin. (Note that since all the norms on a finite dimensional vector space are equivalent, the notions of boundedness and opennes are independent of the norm).

$B$ is open $\Rightarrow$ $\exists y>0$ such that $(0,y)\in B \Rightarrow A^n\begin{pmatrix} 0 \\ y \end{pmatrix}= \begin{pmatrix} ny \\ y \end{pmatrix} \in B$ for every $n \in \mathbb{N}$. This contradicts the boundedness of $B$ (w.r.t to the standard Euclidean norm for instance).

**Last remark:**

If we want to be more restrictive and require $T$ to be an isometry of some inner product, then the answer is quite simple.

$T$ preserves some inner product on $V$ if and only if $V$ admits a basis for which the matrix of $T$ is orthogonal (in other words the matrix of $T$ on an arbitrary basis is similar to an orthogonal matrix). The occurs if and only if the complexification of T is diagonalisable, and all its (complex) eigenvalues have absolute value 1.

**I am interested to know what additional potential isometries we can get when we allow more flexibility.** (That is we allow arbitrary norms, not just those that come from inner products).