There are a few ways to justify it.
First, you could look at the motion of the object as it rotates. In 2D, it turns out that all such motion can be decomposed into motion of the CM, and rotation about the CM. Therefore, rotating around the CM itself is the only way to guarantee the CM doesn't move, and hence is the least energetically costly.
Another way is by direct calculus. Suppose some 1D object is made of point masses at positions $x_i$ with masses $m_i$. Then
$$I(x) = \sum m_i (x_i-x)^2$$
is the moment of inertia about $x$. The minimum is attained when the derivative is zero, so
$$0 = 2 \sum m_i (x_i - x)$$
which implies that
$$x = \frac{\sum m_i x_i}{\sum m_i}.$$
This is the definition of the center of mass.
The definition of the MMOI tensor is to convert angular velocity vector, to angular momentum vector (both vectors on the same basis vectors).
But for a rigid body, the MMOI tensor is defined for body riding basis vector (fixed to the body), such that if you know the rotational velocity vector on the body coordinates you could write
$$ \boldsymbol{L}_{\rm body} = \mathbf{I}_{\rm body\,} \boldsymbol{\omega}_{\rm body} \tag{1}$$
Each of these two vectors can be re-oriented to the common inertial frame (world basis vectors) using the local-to-world 3×3 rotation matrix $\mathbf{R}$
$$ \begin{aligned}
\boldsymbol{\omega} &= \mathbf{R}\,\boldsymbol{\omega}_{\rm body} \\
\boldsymbol{L} &= \mathbf{R}\,\boldsymbol{L}_{\rm body} \\
\end{aligned} \tag{2}$$
Not the inverse rotation transformation is $\mathrm{R}^\top$ so angular momentum in the world frame is
$$ \begin{aligned}
\boldsymbol{L} & = \mathbf{R} \boldsymbol{L}_{\rm body} \\
& = \mathbf{R} \mathbf{I}_{\rm body} \boldsymbol{\omega}_{\rm body} \\
& = \underbrace{\mathbf{R} \mathbf{I}_{\rm body} \mathbf{R}^\top}_{\rm I} \boldsymbol{\omega} \\
\boldsymbol{L} & = \mathbf{I}\, \boldsymbol{\omega}
\end{aligned}$$
So by definition
$$ \mathbf{I} = \mathbf{R} \mathbf{I}_{\rm body} \mathbf{R}^\top$$ is the calculation of MMOI tensor on a general rotated state.
Best Answer
Provided the rotating object is made to rotate around another axis, then yes. You always need an axis of rotation for moment of inertia to make sense.
You can define the moment of inertia about any axis, and not just the axis of rotation of the object, using the parallel axis theorem which states that if a body has a moment of inertia $I$, and if the object now rotates about another axis (parallel to the original axis) then $$I'=I+ml^2$$ where $l$ is the perpendicular distance between the two axes, and $I'$ is the moment of inertia about the new axis.