= Solution
A map $f:\mathbb R^n\to\mathbb R^m$ is <differentiable> at $x$ if there is a <linear map> $L$ such that
$$
f(x+h)=f(x)+Lh+o(\lVert h\rVert);
$$
then $Df|_x=L$. The <inverse function theorem> says that if $f$ is continuously <differentiable> near $x$ and $Df|_x$ is invertible, then $f$ restricts to a $C^1$ diffeomorphism between neighborhoods of $x$ and $f(x)$.
Since $F(A)=A^TA$ is <polynomial>,
$$
DF|_A(H)=H^TA+A^TH.
$$
Thus $\ker DF|_I=T$, the space of skew-symmetric <matrices>.
Define
$$
\Phi(A)=A^TA+A-A^T-I.
$$
Then $D\Phi|_I(H)=2H$, so the inverse <function> theorem supplies open neighborhoods $I\in U$, $0\in V$ on which $\Phi:U\to V$ is a $C^1$ diffeomorphism. Its symmetric part is $A^TA-I$ and its skew part is $A-A^T$. Consequently
$$
\Phi(A)\in T\iff A^TA=I,
$$
and therefore $\Phi(U\cap\mathcal O)=V\cap T$.
For any $R\in\mathcal O$, left multiplication $A\mapsto R^TA$ is a homeomorphism preserving $\mathcal O$ and carrying $R$ to $I$. Transporting the preceding chart gives a neighborhood of $R$ in $\mathcal O$ homeomorphic to an open subset of $T$.
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