TITLE: Proof of the extrinsic to intrinsic rotation transform
QUESTION [8 upvotes]: Wikipedia states that:

Any extrinsic rotation is equivalent to an intrinsic rotation by the
  same angles but with inverted order of elemental rotations, and
  vice-versa. For instance, the intrinsic rotations $x-y’-z''$ by angles $\alpha, \beta,\gamma$ are equivalent to the extrinsic rotations $z-y-x$ by angles $\gamma, \beta, \alpha$. 

Is there a simple proof why this is so?

REPLY [6 votes]: I googled on the Internet, but also found not much material on this topic.
The most relevant references are from Wikipedia as @Apoo suggested, and a blog the proof is at bottom of the page.
Though the arguments are quite complete, neither of them could convince me. In order to prove the statement, I have to first introduce changing a transformation between coordinate systems. There is a derivation from blender stackexchange, and I excerpt the equation as follows:
$$
T_{\text {world}}=S_{\text {world}} \times T_{s} \times S_{\text {world}}^{-1}, \tag{1}
$$
where $T_{\text {world}}$ is the transformation matrix in world coordinate system,
$S_{\text {world}}$ is the world matrix of the referenced object $s$,  $T_{s}$ is the local transformation matrix based on $s
$.
Assume that an intrinsic Euler angle can be represented as a product of three rotation matrices
$$
R = Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha).
$$
The goal is to prove, there exists an extrinsic rotation sequence, s.t.
$$
Z^{\prime \prime}(\gamma) Y^{\prime}(\beta) X(\alpha) =
X(\alpha)Y(\beta)Z(\gamma).
$$
Consider $Y^\prime(\beta)$ as a rotation matrix around $y^\prime$ of angle $\beta$ relative to $x^\prime - y^\prime - z^\prime$ coordinate system, we can obtain the corresponding rotation matrix in $x-y-z$ coordinate system using conversion eq. $(1)$:
$$
Y^\prime = X Y X^{-1},
$$
it follows that
$$
Y =  X^{-1} Y^\prime X . \tag{2}
$$
Through a similar argument, we can also get the world rotation matrix of angle $\gamma$ around $z^{\prime \prime}$ in $x-y-z$ coordinate system:
$$
Z^{\prime \prime} = (Y^\prime X) Z (Y^\prime X)^{-1},
$$
and it follows that
$$
Z = (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X). \tag{3}
$$
Multiplying $X$ by eq. $(2)$ and eq. $(3)$, we obtain
$$
\begin{align}
X Y Z &= X X^{-1} Y^\prime X (Y^\prime X)^{-1} Z^{\prime \prime} (Y^\prime X) \\
      &= Z^{\prime \prime} Y^{\prime} X ,
\end{align}
$$
which completes the proof.