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.