The convex hull of rotations does not contain reflections $\newcommand{\SO}{\operatorname{SO}_n}$
$\newcommand{\Om}{\operatorname{O}_n^{-}}$
I saw here the following claim:
Let $\SO$ be the special orthogonal group, and let $\Om$ be the orthogonal matrices of determinant $-1$. Then the convex hull of $\SO$ does not contain any element from $\Om$.
I tried to prove this "directly", and reduced this to the following claim:
Let $A,B \in \SO$. Suppose that
$$
AB^T+BA^T=2\operatorname{Id}. \tag{1}
$$
Then $A=B$. Writing $R=AB^T$, this reduces to proving that if
$$
R+R^T=2\operatorname{Id}, \tag{2}
$$
then $R=\operatorname{Id}$.
Is there an easy elementary proof of this? I were able to show this directly for $n=2$ using the explicit formula for 2D rotation matrices. Using canonical forms, I can reduce the higher dimensional case to the two-dimensional case.
Is there a proof which avoids using canonical forms?
I can show that if $R$ satisfies equation $(2)$, then so does $R^n$ for every natural $n$, but I am not sure this helps.

I can show $(2)$ implies $R = \operatorname{Id}$, using strict convexity of the norm. We see that, for any $x \in \Bbb{R}^n$,
$$\|Rx + R^\top x + x\| = \|3x\| = \|x\| + \|x\| + \|x\| = \|Rx\| + \|R^\top x\| + \|x\|,$$
hence $Rx$, $R^\top x$, and $x$ are all non-negative scalar multiples of each other. Since all three have the same norm, they must be equal. Thus $x = Rx$. This holds true for all $x$, so $R = \operatorname{Id}$.
