Which special orthogonal groups are ambivalent? A group is ambivalent if every element is conjugate to its inverse.
$SO(1)$, the trivial group, is obviously ambivalent.
$SO(2)\cong \mathbb{R}/\mathbb{Z}$ is not.
$SO(3)$ is, however; any three-dimensional rotation is a rotation about some axis $A$, and viewing the rotation from the other side of $A$ will present the rotation as occurring in the opposite direction (so any element of $SO(3)$ which exchanges the two ends of $A$ suffices for the conjugation). Similar reasoning shows that $A_5$, the symmetry group of the icosahedron, must be ambivalent.
What about other $SO(n)$?
 A: Every element $g\in SO(n)$ admits a normal form, i.e. is conjugate (in $SO(n)$) to a block-diagonal matrix $A$ with $k$ 2-by-2 blocks given by rotational matrices $R_{\phi_i}$ and, possibly, one more (last) 1-by-1 block $[1]$. This, of course, depends on the parity of $n$: In the former case, $n=2k$ and in the latter case $n=2k+1$.
My convention here and below is that $\phi_i\in (-\pi, \pi]$.

*

*Suppose first that $n$ is divisible by $4$, i.e. both $n$ and $k$ are even. Then consider the involution $\tau\in O(n)$ represented by a block-diagonal matrix with the 2-by-2 blocks  equal
$$
\left[\begin{array}{cc}0&1\\
1&0\\
\end{array}\right]. 
$$
This involution satisfies $\tau A \tau = A^{-1}$. Since $k$ is even, $\det(\tau)=1$, i.e. $\tau\in SO(n)$.


*Suppose that $n=2k+1$ is odd. I will use a similar involution $\tau$, except for the last 1-by-1 block I will use $\pm 1$, depending on the parity of $k$, to ensure that $\det(\tau)=1$. Again, $\tau\in O(n)$ and  $\tau A \tau = A^{-1}$.
Together, these observations show that unless $n$ is even and not divisible by $4$, the group $SO(n)$ is ambivalent.


*Lastly, suppose that $n$ is even and not divisible by $4$. I will take a regular (aka generic) orthogonal matrix $A$, where all the angles satisfy $|\phi_i|\ne |\phi_j|$ for all $i\ne j$.

The orthogonal decomposition of ${\mathbb R}^n$ in 2-dimensional $A$-invariant subspaces $V_1,...,V_k$ is then unique and, moreover, if $h\in O(n)$ satisfies $hAh^{-1}=A^{-1}$, then $h$ preserves each $V_k$ and reverses its orientation. In other words, as a matrix, $h$ has block-diagonal form where each block $h_i$ is 2-by-2 and satisfies $\det(h_i)=-1$.
Since
$$
\det(h)=\prod_{i=1}^k \det(h_k)= (-1)^k,
$$
we get $\det(h)=-1$. Hence, $h\notin SO(n)$. Thus, $SO(n)$ is not ambivalent in this case.
