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At least one example of dimension $2$ (but
I don't have an example two, and perhaps infinitely many, examples in other finite dimension)dimension $n$.
n=2. Take $X={\mathbb R}^2$ with $\ell^1$-norm $$\|x\|_1=|x_1|+|x_2|.$$ Then $X^*={\mathbb R}^2$ has the $\ell^\infty$-norm $$\|y\|_\infty=\max(|y_1|,|y_2|).$$ I turns out that $$\|x\|_1=\max(|x_1+x_2|,|x_1-x_2|)$$ and thus $X^*$ X'$is isometric to$X$, via$x\mapsto(x_1+x_2,x_1-x_2)$. More generally, suppose that in$\mathbb R^n$, we have a convex polytope$T$that is self-dual and is symmetric under$x\leftrightarrow-x$. Let$\|\cdot\|_T$be the gauge associated with$T$. Then$X=(\mathbb R^n, \|\cdot\|_T)$is isometric to$X'$because$T$is the unit ball of$X$and$T'=T$is that of$X'$. For instance, if n=4, the polyoctahedron (= octaplex) has these properties, thus there is an$\mathbb R^4$that is isometric to its dual, yet is not Hilbert. If$n\ge3$, the simplex is self-dual but not centro-symmetric. This raises two questions: Does there exist other centro-symmetric self dual convex polytopes? Maybe there exist one in any even dimension ... Is it possible to deform the examples above so as to replace the polygone/-tope by a ball with a smooth boundary? 1 At least one example of dimension$2$(but I don't have an example in other finite dimension). Take$X={\mathbb R}^2$with$\ell^1$-norm $$\|x\|_1=|x_1|+|x_2|.$$ Then$X^*={\mathbb R}^2$has the$\ell^\infty$-norm $$\|y\|_\infty=\max(|y_1|,|y_2|).$$ I turns out that $$\|x\|_1=\max(|x_1+x_2|,|x_1-x_2|)$$ and thus$X^*$is isometric to$X$, via$x\mapsto(x_1+x_2,x_1-x_2)\$.