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Determine the group of isometries of $\mathbb{R}^n$ equipped with the sup metric.

My wild guess is $(a_1,\ldots,a_n)\mapsto(\pm a_{\sigma(1)},\ldots,\pm a_{\sigma(n)})+(c_1,\ldots,c_n)$ where $(c_1,\ldots,c_n)$ is constant, the choice of $\pm$'s are the same for all elements in the domain, and $\sigma$ is a permutation of $(1,2,\ldots,n)$. What I note was that subtracting the function by a constant, changing the sign in one entry, and permuting the entries preserve the isometric property.

So we can assume $f(0,\ldots,0)=(0,\ldots,0)$. For any vector $(a_1,\ldots,a_n)$, $\|f(a_1,\ldots,a_n)-(0,\ldots,0)\|=\|(a_1,\ldots,a_n)-(0,\ldots,0)\|$, so $\|f(a_1,\ldots,a_n)\|=\max|a_i|$. Therefore, for some $k$, there are infinitely many vectors $(a_1,\ldots,a_n)$ with $\|f(a_1,\ldots,a_n)\|=|a_k|$. Among them, there are infinitely many $(a_1,\ldots,a_n)$ such that $\|f(a_1,\ldots,a_n)\|$ is the absolute value of its $j$-th position. By permuting the entries, we can assume $j=k=1$. These vectors have the property that the first entries of $f(a_1,\ldots,a_n)$ and $(a_1,\ldots,a_n)$ coincide in absolute value. Since we can change the sign in one entry, assume infinitely many of them coincide with the same sign.

I was thinking, from here we may be able to prove the same result for any vector, i.e., the first entries of $f(a_1,\ldots,a_n)$ and $(a_1,\ldots,a_n)$ are equal. Then, we can move on to the second, third, all the way up to the $n$-th entry. But I got stucked here.

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Think about what an isometry fixing $0$ has to do when restricted to the extremal points of the unit ball. Your wild guess is correct. –  t.b. Mar 24 '12 at 12:03
I managed to prove that the isometry sends the extremal points into themselves (it's very difficult for me :P). But I cannot prove that it maps "correctly", and I don't know how to continue from here. –  pola Mar 24 '12 at 13:01
Notice that every vector of the standard basis is uniquely a combination of $2^{n-1}$ of the extremal points and conversely, such a combination of norm $1$ of $2^{n-1}$ of the extremal points is $\pm$ a standard basis vector (prove this by induction). –  t.b. Mar 24 '12 at 13:09
By combination I mean of course a convex combination. –  t.b. Mar 24 '12 at 13:24
What is a combination of norm 1? Does it mean the arithmetic mean? Anyway, I haven't been able to prove it, but can you give a direction on how to complete the proof after this? I'm not sure how to make use of it, because the norm function is not linear. –  pola Mar 24 '12 at 13:42
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