Consider an ordered vector space $V$ over $\mathbb{Q}$ in the usual language of ordered vector spaces $(<,0,-,+,\lambda\cdot)_{\lambda \in \mathbb{Q}}$. For the duration of this question, the word interval refers to a non-degenerate bounded open interval within such a vector space.
Call two intervals $(a,b) \subseteq V$ and $(c,d) \subseteq V$ definably isomorphic if we can find an invertible function definable with parameters from $V$ that maps $(a,b)$ to $(c,d)$.
One can easily construct an ordered vector space over $\mathbb{Q}$ containing two intervals that are not definably isomorphic to each other. For example, take the lexicographic product $\mathbb{R} \times \mathbb{Q}$, and compare the uncountable interval between the vectors $(0,0)$ and $(1,0)$ to the countable interval between the vectors $(0,0)$ and $(0,1)$. The two intervals in this example are not equinumerous in the first place: one cannot even have a bijection between them, much less a definable isomorphism.
Question: What's a good example of an ordered vector space structure over $\mathbb{Q}$ containing two equinumerous, but not definably isomorphic intervals? Preferably one that can be presented / explained easily to non-logicians (e.g. ideally not a Henkin construction).
Note: It follows from the usual o-minimality results that such an isomorphism is piecewise affine, with finitely many pieces. One can have a definable isomorphism between two intervals even if one cannot have an affine isomorphism. For example, consider $\mathbb{Q}(\sqrt{2})$ as an ordered vector space over $\mathbb{Q}$. The intervals $(0,\sqrt{2})$ and $(0,1)$ have no affine isomorphism between them since $\sqrt{2}$ and $1$ are linearly independent over $\mathbb{Q}$, but one can still write down a piecewise affine definable isomorphism $f$ between the two intervals as follows.
$$f(x) = \begin{cases} x & \text{ if } x \in (0,2-\sqrt{2})\\ \frac{1}{2}(x - 2 + \sqrt{2}) + 2 - \sqrt{2} & \text{ otherwise} \end{cases}$$
In $\mathbb{Q}^2$ ordered lexicographically, the intervals $((0,0),(0,1))$ and $((0,0),(1,0))$ are both countable, but there is no piecewise affine bijection between them (where piecewise means "with finitely many pieces").
To see this, suppose we have an affine map $A$ given by $$(x,y)\mapsto q(x,y)+(a,b) = (qx+a,qy+b).$$ Then for all $y$, $A(0,y) = (a,qy+b)$, so the first coordinate of the output of $A$ is constant on the entire interval $((0,0),(0,1))$. So a piecewise affine map (with finitely many pieces) only takes on finitely many first-coordinate output values on the interval $((0,0),(0,1))$, and thus it cannot be surjective onto $((0,0),(1,0))$.