Disconnected closed subgroups of $SO(n)$

If $G\le\mathrm{SO}(m)$ is closed and connected, and $K\le\mathrm{SO}(n)$ is finite, then $G\times K\le\mathrm{SO}(m+n)$ fits the bill. The identity component will be $G$, which has a complementary subgroup $K$ in $H=G\times K$, but in general this isn't true. For instance, consider

$$H=\mathrm{SO}(n)\sqcup i[\mathrm{O}(n)\setminus\mathrm{SO}(n)]\le \mathrm{U}(n)$$

embedded in $\mathrm{SO}(2n)$ when $n$ is odd; here $\mathrm{SO}(n)$ has no complementary subgroup in $H$ - exercise. (Also we can give an alternate description of $H$ with Kronecker products instead of complex numbers if you want.)

In general though, the identity component $H_0$ will be connected and $H\le N_G(H_0)$, so maybe we can cook up a general recipe based on the idea of looking for nontrivial finite subgroups of $N_G(H_0)/H_0$ for closed, connected subgroups $H_0\le\mathrm{SO}(n)$.