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I would like to find an example for two norms on $\| \cdot \|_i$ on $\mathbb{R}^n$ ($i=1,2$) such that both unit spheres has the same finite number of extreme points, but $\| \cdot \|_1,\| \cdot \|_2$ are not isometric.

Are there easy explicit examples?

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    Do you know whether any isometry between such spaces must be linear? Because it seems to be true when both are strictly convex, but this is not the case here – 2017-02-21

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For more fun, let's do it in 3 dimensions. Take for the unit ball of the first space the regular icosahedron (12 vertices). For the second space take the unit ball to be a double decagonal pyramid: say the vertices are 10 equally spaced points around the equator, together with the north and south poles. Clearly any linear isomorphism preserves co-planarity, but no plane contains 10 of the 12 vertices of the icosahedron.

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    Dear GEdgar, I have a question: does linearity follow from the assumptions or you are assuming it? Strict convexity of both norms would imply it for instance (but this is not the case here) – 2017-02-22
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    @Del ... I assumed that linearity was included when Asaf said "isometric" for normed spaces. But if not, use the Mazur-Ulam theorem https://en.wikipedia.org/wiki/Mazur%2DUlam%5Ftheorem – 2017-02-22
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    Many thanks! I didn't know this theorem. Exactly what I was looking (and hoping) for – 2017-02-22
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You can define a norm on $\mathbf{R}^2$ by picking any balanced convex hexagon to be the unit sphere. It's enough to pick two such hexagons which are not images of one another under any linear mapping. This guarantees there is no (linear) isometry between your two spaces.

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    These spaces are isometric via the map $f(x,y)=(x+y,x-y)$. – 2017-02-20
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    @AsafShachar Thanks. – 2017-02-20
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    I've fixed the answer, I hope. – 2017-02-21