The statement you make about tetrahedra generalizes to arbitrary $n$. Specifically, the symmetric group $S_n$ is the group of symmetries of a regular $n$-simplex, and the alternating group $A_n$ acts on this simplex by rotations. (In fact, $A_n$ is precisely the set of rotational symmetries.) Elements of different conjugacy classes of $A_n$ are geometrically distinguishable, in the sense that they would "look different" to an observer in $\mathbb{R}^n$.
One way of making the notion of "look different" precise is that non-conjugate elements of $A_n$ correspond to non-conjugate elements of the rotation group $SO(n)$. Thus, two non-conjugate elements of $A_n$ do not look the same up to rotation of the simplex.
Incidentally, the simplest algorithm to distinguish conjugacy classes in $A_n$ is essentially to check the sign of the conjugator. For example, the elements $(5)(2\;6\;3)(1\;9\;4\;8\;7)$ and $(2)(1\;4\;8)(3\;7\;5\;6\;9)$ are conjugate in $S_9$, and their conjugacy class in $S_9$ splits into two conjugacy classes in $A_9$. To check whether the two elements are conjugate in $A_9$, we consider a permutation that maps between corresponding numbers:
$$
\begin{bmatrix}
5 & 2 & 6 & 3 & 1 & 9 & 4 & 8 & 7\\
2 & 1 & 4 & 8 & 3 & 7 & 5 & 6 & 9
\end{bmatrix} \;=\; (1\;3\;8\;6\;4\;5\;2)(7\;9)
$$
This permutation is odd, so the two elements are not in the same conjugacy class in $A_9$. It is possible to construe this algorithm geometrically, since the difference between odd and even permutations is the same as the difference between right-handed and left-handed coordinate systems.
Best Answer
Background
By computing the determinant of the defining condition $AA^T=I$ of the group $O(2)$ we see that the determinant of an orthogonal matrix is either $+1$ or $-1$. This shows that $SO(2)$ is a subgroup of $O(2)$ with index two.
In two dimensions, we can use the condition that the columns of an orthogonal matrix must be unit norm orthogonal vectors, we can obtain the following parametrization of orthogonal matrices with determinant $+1$
$$ R(\theta)=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} $$
and the following parametrization of orthogonal matrices with determinant $-1$
$$ R(\theta)Z=\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix} $$
where $Z=\mathrm{diag}(1, -1)$. The matrices of the form $R(\theta)$ describe rotations and the matrices of the form $R(\theta)Z$ describe rotoreflections.
Note that $R(\theta)Z=ZR(-\theta)$ for any angle $\theta$.
Conjugacy classes
Conjugacy class $\mathrm{Cl}(R(\alpha))$ of a rotation $R(\alpha)$ is
$$ \begin{align} \mathrm{Cl}(R(\alpha))&=\{R(\theta)R(\alpha)R(-\theta)\,|\,\theta\in[0, 2\pi)\}\cup\{R(\theta)ZR(\alpha)ZR(-\theta)\,|\,\theta\in[0, 2\pi)\} \\ &= \{R(\alpha)\}\cup\{R(-\alpha)\} \\ &= \{R(\alpha),R(-\alpha)\}. \end{align} $$
Conjugacy class of a rotoreflection $R(\alpha)Z$ is
$$ \begin{align} \mathrm{Cl}(R(\alpha)Z)&=\{R(\theta)R(\alpha)ZR(-\theta)\,|\,\theta\in[0, 2\pi)\}\cup\{R(\theta)ZR(\alpha)ZZR(-\theta)\,|\,\theta\in[0, 2\pi)\} \\ &= \{R(\theta)Z\,|\,\theta\in[0,2\pi)\}\cup\{R(\theta)Z\,|\,\theta\in[0,2\pi)\} \\ &= \{R(\theta)Z\,|\,\theta\in[0,2\pi)\}. \end{align} $$
We conclude that every rotation is conjugate to itself and its inverse and all rotoreflections are conjugate to each other.