This is ex. from Munkers book: \ find a linear transformation $h: \mathbb{R}^n \rightarrow \mathbb{R}^n$ that preserves volumes but is not an isometry.
It's clear that $n$ should be greater than 1, but even in case $n=2$ I'm not able to provide an example.
On
The condition that $h: \Bbb R^n \to \Bbb R^n$ preserve volumes is $\det h = 1$, whereas the condition that it is an isometry is $h^{\top} h = {\bf 1}$. We can generate simple examples as follows: If we identify linear transformations $\Bbb R^n \to \Bbb R^n$ with their matrices w.r.t. the standard basis, we see, for example, for diagonal linear transformations $$h = \pmatrix{\lambda_1 & \cdot \\ \cdot & \lambda_2} ,$$ the volume-preserving condition $\det h = 1$ is $$\lambda_1 \lambda_2 = 1,$$ and the (non-)isometry condition $h^{\top} h \neq {\bf 1}$ is $$\pmatrix{\lambda_1^2 & \cdot \\ \cdot & \lambda_2^2} \neq \pmatrix{1&\cdot\\\cdot&1}.$$ The volume-preserving condition implies $\lambda_2 = \lambda_1^{-1}$, and then the (non-)isometry condition implies $\lambda_1 \not\in \{\pm 1\},$ leaving exactly the examples $$ h = \pmatrix{\lambda & \cdot \\ \cdot & \lambda^{-1}}, \qquad \lambda \not\in \{\pm 1\} . $$
Note that $\det h = 1$ imposes one polynomial condition on the entries of $h$, whereas $h^{\top} h = {\bf 1}$ imposes $\frac{1}{2} n (n + 1)$ polynomial conditions, which is $> 1$ for $n > 1$. Thus, for $n > 1$ we expect a generic volume-preserving transformation is not an isometry, which turns out to be true in a way that can be made precise.
I recall you that a linear transformation $h$ preserves volumes if and only if $|\det h|=1$. So, the simplest example is $$h(x_1 , \dots, x_n)=(2x_1 , x_2 / 2 , x_3 , \dots , x_n)$$ represented by the diagonal matrix whose diagonal elements are $2, 1/2, 1, \dots , 1$.