The diameter of the orthogonal group

Idea. Two points of a "sphere" of radius $r$ in a normed space are at distance at most $2r$, with equality for antipodal points. $O(n)$ lies on the operator-norm sphere of radius $1$ and on the Frobenius sphere of radius $\sqrt n$, and contains the antipodal pair $\pm I$.

Step 1: norms of orthogonal matrices. For $U\in O(n)$, $|Ux|=|x|$ for all $x$, so $\|U\|=1$. And $\|U\|_F^2=\operatorname{tr}(U^TU)=\operatorname{tr}I=n$.

Step 2: upper bounds. For $U,V\in O(n)$,
$$\|U-V\|\le\|U\|+\|V\|=2,\qquad\|U-V\|_F\le\|U\|_F+\|V\|_F=2\sqrt n.$$

Step 3: attainment. $-I$ is orthogonal since $(-I)^T(-I)=I$. Then $\|I-(-I)\|=\|2I\|=2$ and $\|I-(-I)\|_F=2\|I\|_F=2\sqrt n$.

Conclusion. The two diameters are exactly $2$ and $2\sqrt n$.

Remark. For $n=1$, $O(1)=\{\pm1\}$ and the diameter is $2$. Note that $-I$ lies in $SO(n)$ iff $n$ is even. For odd $n\ge3$ the operator-norm diameter of $SO(n)$ is still $2$: the matrices $I$ and $D=\operatorname{diag}(-1,-1,1,\dots,1)\in SO(n)$ satisfy $\|I-D\|=\|\operatorname{diag}(2,2,0,\dots,0)\|=2$, and the upper bound $2$ is the same as above. (For the Frobenius norm the value $2\sqrt n$ is no longer attained when $n$ is odd; we do not need this.)