$$\newcommand{\uvec}[1]{\underline{#1}}.$$
You must be careful. You have to distinguish between vectors and their components. For the vectors the formula reads
$$\dot{\vec{J}} = \vec{\tau}.$$
For the components in the space-fixed (inertial) frame the equation reads the same
$$\dot{\underline{J}}=\underline{\tau},$$
where the underlined symbol are column vectors with the space-fixed components of the corresponding vector.
The body-fixed frame is a rotating frame and thus the time derivative of a vector translates into the formula
$$\dot{\underline{J}}' + \underline{\omega}' \times \underline{J}=\underline{\tau}',$$
where an underlined symbol with prime means column vectors with the body-fixed components of the corresponding vector.
Both the angular momentum and the torque depend on the body-fixed origin, the angular velocity does not. I was wrong about this in #13!
The most general motion of a rigid body is described by first choosing an inertial reference frame ##(O,\vec{e}_j)## (##j \in \{1,2,3 \}##), where ##O## is the origin of this "space-fixed" inertial reference frame and the ##\vec{e}_j## are an arbitrary righthanded Cartesian basis. Then you have to also define a "body-fixed" reference frame, which usually is a non-inertial rotating frame. To that end you define an arbitrary point ##O'## fixed relative to the body and a righthanded body-fixed Cartesian basis ##\vec{e}_k'##.
Then any point ##P_k## of the rigid body (which for simplicity we assume to consist of a discrete set of point masses) is described by the position vector ##\vec{x}_k=\overrightarrow{OP_k}##. Now we write
$$\vec{x}_k=\vec{R} + \vec{r}_k \quad \text{with} \quad \vec{R}=\overrightarrow{OO'}, \quad \vec{r}=\overrightarrow{O'P_k}.$$
These are all basis-independent vectors.
A rigid body is now by definition a body, where for all of its points ##P_k##
$$\uvec{r}_k'=\text{const}.$$
Thus for body-fixed relative vectors ##\vec{r}_k## it's convenient to use body-fixed components, which are time-independent.
Now for any vector ##\vec{V}## you have, introducing the transformation matrix between the space-fixed and the body-fixed Cartesian bases (Cartesian Einstein summation convention used):
$$\vec{e}_k'=\vec{e}_j D_{jk},$$
which is ##(D_{jk})\in \mathrm{SO}(3)##, i.e., it's a rotation matrix fulfilling ##\hat{D}^{\text{T}}=\hat{D}^{-1}## and ##\mathrm{det} \hat{D}=+1##,
$$\vec{V}=V_k' \vec{e}_k'=\vec{e}_j V_k' D_{jk} = \vec{e}_j V_j \; \Rightarrow \; \uvec{V}'=\hat{D} \uvec{V}.$$
Then on the one hand
$$\dot{\vec{V}} = \vec{e}_j \dot{V}_j,$$
because by definition the space-fixed basis is time independent but on the other hand
$$\dot{\vec{V}}=\dot{V}_k' \vec{e}_k' + V_{k}' \dot{\vec{e}}_k',$$
because the body-fixed basis is time-dependent.
To get the body-fixed components of ##\dot{\vec{V}}## it's more convenient to use
$$\uvec{V}=\hat{D}^{\text{T}} \uvec{V}' \; \Rightarrow \; \dot{\uvec{V}}=\hat{D}^{\text{T}} \dot{\uvec{V}}' + \dot{\hat{D}}^{\text{T}} \uvec{V}'.$$
Thus the body-fixed components of ##\dot{\vec{V}}##, which I denote as ##\mathrm{D}_t \uvec{V}'##, are given by
$$\mathrm{D}_t \uvec{V}'=\hat{D} \dot{\uvec{V}}=\dot{\uvec{V}}' + \hat{D} \dot{\hat{D}}^{\text{T}} \uvec{V}'.$$
Now the matrix in the 2nd term is antisymmetric, because
$$\hat{D} \hat{D}^{\text{T}}=\hat{1} \; \rightarrow \; \mathrm{d}_t (\hat{D} \hat{D}^{\text{T}}) = \dot{\hat{D}} \hat{D}^{\text{T}} + \hat{D} \dot{\hat{D}}^{\text{T}}=0$$
and from that
$$\hat{\Omega}'=\hat{D} \dot{\hat{D}}^{\text{T}}=-\dot{\hat{D}} \hat{D}^{\text{T}}=-\hat{\Omega}^{\prime \text{T}}.$$
Thus one can write
$$\Omega_{kl}'=-\epsilon_{klm} \omega_m',$$
where
$$\omega_m'=-\frac{1}{2} \epsilon_{klm} \Omega_{kl}$$
is the Hodge dual of ##\Omega_{kl}'## (which is a one-to-one mapping between antisymmetric matrices and vector components). Thus we finally have
$$(\mathrm{D}_t \uvec{V}')_k=\dot{V}_k' + \hat{\Omega}_{kl}' V_l'=\dot{V}_k' - \epsilon_{klm} \omega_m' V_l' = \dot{V}_k'+\epsilon_{kml} \omega_m' V_l',$$
i.e., in column-vector notation
$$\mathrm{D}_t \uvec{V}'=\dot{\uvec{V}}' + \uvec{\omega}' \times \uvec{V}'.$$
Supposed you have a spinning top, i.e., if you fix the body at one of its points (to make the thing simpler) and choose ##O=O'##, i.e., you make the origin of the space-fixed reference the same as the origin of the body-fixed origin ##O'## which you choose as the point of the body which is fixed, then you have
$$\mathrm{D}_t \uvec{J}' = \dot{\uvec{J}}'+\uvec{\omega}' \times \uvec{V}' = \uvec{\tau}'.$$
This is an equation for body-fixed components, which is convenient, because as the further analysis of this case shows there's a symmetric tensor with time-independent body-fixed components ##\uvec{\Theta}'## such that
$$\uvec{J}'=\hat{\Theta}' \uvec{\omega}',$$
there ##\overleftrightarrow{\Theta}=\Theta_{kl}' \vec{e}_k' \otimes \vec{e}_l'## is the tensor of inertia. Since it's symmetric, it can be diagonalized by choosing the body-fixed basis as oriented along the corresponding principle axes, i.e., you make the body-fixed basis vectors the eigenvectors of the symmetric matrix. A well-known theorem tells you that you can always make these eigenvectors to form a right-handed Cartesian basis, so that the above considerations apply to this most convenient choice of the body-fixed basis. With ##\hat{\Theta}'=\mathrm{diag}(A,B,C)## you have
$$\uvec{J}'=\begin{pmatrix} J_1' \\ J_2' \\ J_3' \end{pmatrix} = \begin{pmatrix} A \omega_1' \\ B \omega_2' \\ C \omega_3' \end{pmatrix},$$
and plugging this in the above equation for ##\mathrm{D}_t \uvec{J}'## you get Euler's equation of motion for ##\uvec{\omega}'## of the spinning top.
See
https://en.wikipedia.org/wiki/Euler's_equations_(rigid_body_dynamics)
where they use of course somewhat different notation.
! I wonder if I'm understanding this correctly.
