Here is a natural map $F\colon O(n)\to S_n$ which does not have the specified properties, because I did not read them correctly at first. Nonetheless, it may be of interest, if only as an example of something to avoid.

The idea is to use the Bruhat decomposition $GL_n(\mathbb{R})=\coprod_{\sigma\in S_n}B\sigma B$.

In more detail, consider two flags $0=U_0<U_1<\dotsb<U_n=\mathbb{R^n}$ and $0=V_0<V_1<\dotsb<V_n=\mathbb{R^n}$. For $0\leq i,j<n$ put $$ Q_{ij} = \frac{U_i\cap V_j}{(U_i\cap V_{j-1})+(U_{i-1}\cap V_j)} \simeq \frac{U_{i-1}+(U_i\cap V_j)}{U_{i-1}+(U_i\cap V_{j-1})} \simeq \frac{(U_i\cap V_j)+V_{j-1}}{(U_{i-1}\cap V_j)+V_{j-1}} $$ One can check that there is a unique permutation $\sigma$ such that $Q_{ij}=0$ unless $i=\sigma(j)$ in which case $Q_{ij}\simeq\mathbb{R}$. (Indeed, the third description of $Q_{ij}$ makes it clear that there is a unique function $\sigma$ with this property, the second description makes it clear that there is a unique function $\tau$ such that $Q_{ij}=0$ unless $j=\tau(i)$, and then we conclude that $\sigma$ and $\tau$ are inverse to each other and so must be permutations.)

Now consider an element $g\in O(n)$. Let $U_i$ be the span of $e_1,\dotsc,e_i$, and let $V_i$ be the span of $ge_1,\dotsc,ge_i$. Let $F(g)$ be the permutation corresponding to this pair of flags. This is a fairly natural map $O(n)\to S_n$ that is the identity on $S_n$. Everything works in essentially the same way over $\mathbb{C}$.

Here is a natural map $F\colon O(n)\to S_n$ which does not have the specified properties, because I did not read them correctly at first. Nonetheless, it may be of interest, if only as an example of something to avoid.

The idea is to use the Bruhat decomposition $GL_n(\mathbb{R})=\coprod_{\sigma\in S_n}B\sigma B$.

In more detail, consider two flags $0=U_0<U_1<\dotsb<U_n=\mathbb{R}^n$ and $0=V_0<V_1<\dotsb<V_n=\mathbb{R}^n$. For $0\leq i,j<n$ put $$ Q_{ij} = \frac{U_i\cap V_j}{(U_i\cap V_{j-1})+(U_{i-1}\cap V_j)} \simeq \frac{U_{i-1}+(U_i\cap V_j)}{U_{i-1}+(U_i\cap V_{j-1})} \simeq \frac{(U_i\cap V_j)+V_{j-1}}{(U_{i-1}\cap V_j)+V_{j-1}} $$ One can check that there is a unique permutation $\sigma$ such that $Q_{ij}=0$ unless $i=\sigma(j)$ in which case $Q_{ij}\simeq\mathbb{R}$. (Indeed, the third description of $Q_{ij}$ makes it clear that there is a unique function $\sigma$ with this property, the second description makes it clear that there is a unique function $\tau$ such that $Q_{ij}=0$ unless $j=\tau(i)$, and then we conclude that $\sigma$ and $\tau$ are inverse to each other and so must be permutations.)

Now consider an element $g\in O(n)$. Let $U_i$ be the span of $e_1,\dotsc,e_i$, and let $V_i$ be the span of $ge_1,\dotsc,ge_i$. Let $F(g)$ be the permutation corresponding to this pair of flags. This is a fairly natural map $O(n)\to S_n$ that is the identity on $S_n$. Everything works in essentially the same way over $\mathbb{C}$.