Short answer
For sure, the components of the angular velocity w.r.t. the body frame, moving with the body, can change in time during the motion of a rigid body. The evolution of the components of the angular velocity w.r.t. the body frame are governed by the Euler equations (nothing more than the components in th\dfrac{d \boldsymbol{\hat{\omega}}}{dt}e body frame of the dynamical equation for the rotational dynamics), that can be written as
$I_{G,XX} \dfrac{d \Omega_X}{dt} - ( I_{G,YY} - I_{G,ZZ} ) \Omega_Y \Omega_Z = M_{G,X}^{ext}$
$I_{G,YY} \dfrac{d \Omega_Y}{dt} - ( I_{G,ZZ} - I_{G,XX} ) \Omega_X \Omega_Z = M_{G,Y}^{ext}$
$I_{G,ZZ} \dfrac{d \Omega_Z}{dt} - ( I_{G,XX} - I_{G,YY} ) \Omega_X \Omega_Y = M_{G,Z}^{ext}$ ,
when referred to the center of gravity $G$ using the principal axis of inertia, so that the tensor of inertia is diagonal when written in the the body reference frame
$\mathbb{I}_G = I_{G,XX} \mathbf{\hat{X}}\otimes \mathbf{\hat{X}} + I_{G,YY} \mathbf{\hat{Y}}\otimes \mathbf{\hat{Y}} + I_{G,ZZ} \mathbf{\hat{Z}}\otimes \mathbf{\hat{Z}}$
and the angular velocity is written as $\mathbf{\omega} = \Omega_X \mathbf{\hat{X}} + \Omega_Y \mathbf{\hat{Y}} + \Omega_Z \mathbf{\hat{Z}} = \Omega_i \mathbf{\hat{X}}_i$.
Details
Vector equations.
Equations of dynamics are vector equations, and thus it's convenient to write them as vector/tensor equations using abstract notation. Namely, if we trust in classical mechanics, the dynamical equation for the rotation of a rigid body can be written as
$\dfrac{d \mathbf{\Gamma}_H}{dt} = -\dot{\mathbf{x}}_H \times \mathbf{Q} + \mathbf{M}_H^{ext}$,
being $H$ the pole w.r.t. we evaluate the external moments $\mathbf{M}^{ext}_H$ and the angular momentum $\mathbf{\Gamma}_H$ that can be written as $\mathbf{\Gamma}_H = \mathbb{I}_G \cdot \mathbf{\omega} + (\mathbf{r}_G - \mathbf{r}_H) \times \mathbf{Q}$, and $\mathbf{Q}$ is the momentum of the body.
The form of this equation becomes easier if we use the center of gravity $G$ as the pole $H \equiv G$, namely
$\dfrac{d \mathbf{\Gamma}_G}{dt} = \mathbf{M}_G^{ext} \qquad $ with $\qquad \mathbf{\Gamma}_G = \mathbb{I_G} \cdot \mathbf{\omega}$.
Fixed (inertial) reference frame and body reference frame. Now we can introduce two reference frames with unit-length orthogonal vectors of the basis, usually used in the study of rigid body dynamics:
- the fixed inertial reference frame as $\{\mathbf{x}_i\}_{i=1:3}$ that we assume fixed in space so that their time derivative is identically zero;
- the body reference frame, moving in space with the rigid body, $\{\mathbf{X}_j\}_{j=1:3}$, useful because many properties of a rigid body (as an example, the components of the tensor of inertia) are constant in time when referred to the body reference frame. The angular velocity of the reference frame, is thus the same angular velocity $\mathbf{\omega}$ of the rigid body.
We can use these two reference frames to write every vector and tensor quantity; as an example, a vector $\mathbf{v}$ can be written as
$\mathbf{v} = v_i \mathbf{\hat{x}}_i = V_j \mathbf{\hat{X}}_j$
while its time derivative reads
$\dfrac{d\mathbf{v}}{dt} \quad = \quad \dfrac{d v_i}{dt} \mathbf{\hat{x}}_i + v_i \underbrace{\dfrac{d\mathbf{\hat{x}}_i}{dt}}_{=0} \quad = \quad \dfrac{d V_j}{dt} \mathbf{\hat{X}}_j + V_j \underbrace{\dfrac{d\mathbf{\hat{X}}_j }{dt}}_{=\mathbf{\omega} \times \mathbf{\hat{X}}_j \\ \text{(Poinsot)}} $
$\qquad \quad = \quad \dfrac{d v_i}{dt} \mathbf{\hat{x}}_i \qquad \qquad \quad = \quad \dfrac{d V_j}{dt} \mathbf{\hat{X}}_j + \mathbf{\omega} \times \mathbf{v}$
The angular velocity behaves in a "special way" when its time derivative is evaluated, since $\mathbf{\omega} \times \mathbf{\omega} = \mathbf{0}$. The angular acceleration $\mathbf{\alpha} = \frac{d \mathbf{\omega}}{dt}$ becomes
$\dfrac{d\mathbf{\omega}}{dt} = \dfrac{d \omega_i}{dt} \mathbf{\hat{x}}_i = \dfrac{d \Omega_j}{dt} \mathbf{\hat{X}}_j$$\qquad \qquad$$\mathbf{\alpha} = \alpha_i \mathbf{\hat{x}}_i = A_j \mathbf{\hat{X}}_j$
Change in direction of the angular velocity and orthogonal projectors. If we write the angular velocity $\boldsymbol{\omega} = \omega \boldsymbol{\hat{\omega}}$, its time derivative reads
$\dfrac{d \boldsymbol{\omega}}{dt} = \dfrac{d}{dt} (\omega \boldsymbol{\hat{\omega}}) = \dfrac{d \omega}{dt} \boldsymbol{\hat{\omega}} + \omega \dfrac{d \boldsymbol{\hat{\omega}}}{dt}$.
Since $|\boldsymbol{\hat{\omega}}| = 1 = \text{const.}$, we get that $\frac{d \boldsymbol{\hat{\omega}}}{dt }\cdot \boldsymbol{\hat{\omega}} = 0$, and multiplying the time derivative of $\boldsymbol{\omega}$ by the $\boldsymbol{\hat{\omega}}$,
$\boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt} = \dfrac{d \omega}{dt} \underbrace{\boldsymbol{\hat{\omega}} \cdot \boldsymbol{\hat{\omega}}}_{=1} + \omega \underbrace{\boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\hat{\omega}}}{dt}}_{=0}$$\qquad \rightarrow \qquad$$\dfrac{d \omega}{dt} = \boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$
and using this expression of the time derivative of the magnitude of the angular velocity, we can get the expression of the time derivative of the unit-vector $\boldsymbol{\hat{\omega}}$
$\dfrac{d \boldsymbol{\hat{\omega}}}{dt} = \dfrac{1}{\omega} \left[ \dfrac{d\boldsymbol{\omega}}{dt} - \boldsymbol{\hat{\omega}}\boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}\right] = \dfrac{1}{\omega} \left[ \mathbb{I} - \boldsymbol{\hat{\omega}} \otimes\boldsymbol{\hat{\omega}} \right] \cdot \dfrac{d \boldsymbol{\omega}}{dt} = \dfrac{1}{\omega}\mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$
having introduced the definition of the normal projector $\mathbb{P}_{\perp,\boldsymbol{\omega}} = \mathbb{I} - \boldsymbol{\hat{\omega}} \otimes\boldsymbol{\hat{\omega}}$ in the direction orthogonal to $\boldsymbol{\hat{\omega}}$. Thus, we can summarize that:
time derivative of the magnitude of the angular velocity is the projection of the time derivative of the angular velocity along the angular velocity direction:
$\dfrac{d \omega}{dt} = \boldsymbol{\hat{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$
time derivative of the unit vector indicating the direction of the angular velocity comes from the projection of the time derivative of the angular velocity in the directions orthogonal w.r.t. its direction
$\dfrac{d \boldsymbol{\hat{\omega}}}{dt} = \dfrac{1}{\omega}\mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$
and thus write
$\dfrac{d \boldsymbol{\omega}}{dt} = \underbrace{\dfrac{d \omega}{dt} \boldsymbol{\hat{\omega}}}_{=\mathbb{P}_{//,\boldsymbol{\omega}} \cdot \frac{d \boldsymbol{\omega}}{dt}} + \underbrace{\omega \dfrac{d \boldsymbol{\hat{\omega}}}{dt}}_{=\mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \frac{d \boldsymbol{\omega}}{dt}} = \mathbb{P}_{//,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt} + \mathbb{P}_{\perp,\boldsymbol{\omega}} \cdot \dfrac{d \boldsymbol{\omega}}{dt}$
