In 3 dimensions, a rotation can be characterized by a roll (an $\alpha$ degree rotation around the $x$ axis), pitch (a $\beta$ degree rotation around the $y$ axis), and a yaw (a $\gamma$ degree rotation around the $z$ axis), say in that order. (Let's call roll, pitch, and yaw the different 'modes' of rotation.) Therefore we can express the total rotation $R = R_\alpha * P_\beta * Y_\gamma$ where each is a rotation matrix that I don't want to write out here. We can bash out what this matrix product evaluates to (it's not that difficult), and we find that its trace is $tr(R) = \cos{\alpha} \cos{\beta} + \cos{\beta} \cos{\gamma} + \cos{\gamma}\cos{\alpha} + \sin{\alpha} \sin{\beta} \sin{\gamma}$.
Alternatively, we can characterize $R$ by an axis-of-rotation which it leaves fixed and the angle that everything else gets rotated by (around the axis-of-rotation). This is Euler's Rotation Theorem. Therefore, if we let $T$ be any rotation (there are many) that brings the axis-of-rotation to the x-y axis, and then $S_\theta$ a rotation around the x-y axis by the angle theta (ie. the matrix $\begin{pmatrix}\cos{\theta} & -\sin{\theta} & 0 \\ \sin{\theta} & \cos{\theta} & 0 \\ 0 & 0 & 1\end{pmatrix}$), we can write $R = T S_\theta T^{-1}$. Taking the trace, we get $tr(R) = tr(T S_\theta T^{-1}) = tr(T^{-1} T S_\theta) = tr(S_\theta) = 1 + 2\cos{\theta}$ where we use the following trace property:
Cyclic Permutation Property of Trace (or just the "Trace Property"). For three square matrices $A,B,C$, we have $tr(ABC) = tr(CAB) = tr(BCA) := x$. (Thus we also have $tr(BAC) = tr(ACB) = tr(CBA) :=y$; but $x \ne y$ in general.) Note that $CAB, BCA$ are cyclic permutations of $ABC$, hence the name of the property.
Therefore, we have $1 + 2\cos\theta = \cos{\alpha} \cos{\beta} + \cos{\beta} \cos{\gamma} + \cos{\gamma}\cos{\alpha} + \sin{\alpha} \sin{\beta} \sin{\gamma}$. It is interesting that this expression is symmetric in the variables $\alpha, \beta, \gamma$.
This means that if we exchange the angles of in any of the 3!=6 ways (while keeping the modes in the same order), we obtain rotations that still have the same angle of rotation (but possibly different axes). Call this the "Angle Permutation Property". The Trace Property above also implies the same with the 3 cyclic permutations of the modes of rotation – call that the "Mode Cyclic Permutation Property". Combining the 6 permutations of the angles ($\alpha, \beta, \gamma$) with the 3 cyclic permutations of the modes, we have found a set of 18 related rotations all of which have the same angle of rotation (but possibly different axes).
Is there a geometric argument for the truth of the Angle Permutation Property? of the Mode Cyclic Permutation Property (for which a geometric argument for the Trace Property suffices)? Are their axes related in a similar way as well? What is the geometric significance of this equivalence of these 18 rotations?
By the way, assuming $\alpha, \beta, \gamma$ are all small, we obtain $\theta^2 \approx \alpha^2 + \beta^2 + \gamma^2 (+\alpha\beta\gamma)$, which perhaps makes sense because the three rotations are in some vague sense "orthogonal" (which sense?) but is perhaps surprising given that the surface of the sphere has only two dimensions, not three!
Best Answer
If $A$ is a rotation by angle $θ$ about $\vec v$, and $B$ is a rotation, then the conjugate $BAB^{-1}$ is a rotation by the same angle $θ$ about $B\vec v$. This one fact lets us establish both properties without any trigonometry.
Mode Cyclic Permutation Property
Let $C = \left[\begin{smallmatrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{smallmatrix}\right]$ be the rotation by $\frac{2π}{3}$ about $(1, 1, 1)$. Then $R_α P_β Y_γ$ has the same angle as
\begin{gather*} C(R_αP_βY_γ)C^{-1} = (CR_αC^{-1})(CP_βC^{-1})(CY_γC^{-1}) = P_αY_βR_γ, \\ C^{-1}(R_αP_βY_γ)C = (C^{-1}R_αC)(C^{-1}P_βC)(C^{-1}Y_γC) = Y_αR_βP_γ. \end{gather*}
Angle Permutation Property
$R_αP_βY_γ$ has the same angle as $R_{-α}(R_αP_βY_γ)R_α = P_βY_γR_α$, which has the same angle as $R_βP_γY_α$ by MCPP.
Also, $R_αP_βY_γ$ has the same angle as
\begin{gather*} P_π(R_αP_βY_γ)P_π = (P_πR_αP_π)(P_πP_βP_π)(P_πY_γP_π) = R_{-α}P_βY_{-γ}, \\ Y_{π/2}(R_{-α}P_βY_{-γ})Y_{-π/2} = (Y_{π/2}R_{-α}Y_{-π/2})(Y_{π/2}P_βY_{-π/2})(Y_{π/2}Y_{-γ}Y_{-π/2}) = P_{-α}R_{-β}Y_{-γ}, \end{gather*}
which is the inverse of (thus has the same angle as) $Y_γR_βP_α$, which has the same angle as $R_γP_βY_α$ by MCPP.
Compositions of these two angle permutations $(αβγ)$ and $(αγ)$ generate all six.
Rotation axes
We can extract the corresponding rotation axes from these proofs. If $R_αP_βY_γ$ has axis $\vec v$, then
The axes for all eighteen compositions generated by these will be the six coordinate permutations of $\vec v = (a, b, c)$, the six coordinate permutations of $R_{-a}\vec v = (a, e, f)$, and the six coordinate permutations of $Y_γ\vec v = (d, e, c)$: