Inertia matrix in skew coordinates One has, let's say, the inertia matrix of an object about it's center of gravity: $I_p$ in the coordinate system ($i, j, k$). What is the inertia matrix in another coordinate system, say ($i_1,j_1, k_1$) with $$ \begin{bmatrix} i \\ j\\ k\end{bmatrix} = A \cdot \begin{bmatrix} i_1 \\ j_1 \\ k_1\end{bmatrix}$$ 
 A: You are given $\mathbf A$ which a matrix describing the linear transformation. This matrix will allow you to move from the new coordinates (subscript 1) to the old coordinates (no subscript).
What you have to do now is to transform the inertia matrix $\mathbf I$ into the new basis $\mathbf I_1$. Here is how to come up with the transformation rule: The rotational energy is a scalar (just one number), therefore it has to be the same in every coordinate system. This is
$$ E = \frac12 \vec \omega^\mathrm T \mathbf I \vec \omega \,. $$
You know that $\mathbf A$ will bring you from the old coordinates to the new coordinates. You obtain the new coordinates from the old ones by using $\mathbf A^{-1}$. So $\mathbf A^{-1} \vec \omega$ will be the angular rotational velocity vector in the new coordinate system.
Use the above equation and insert $\mathbf A \mathbf A^{-1}$ into the energy equation above twice: between $\vec \omega^\mathrm T$ and $\mathbf I$ as well as $\mathbf I$ $\vec \omega$. Identify the angular velocity vectors in the new system and the new matrix in between.
What do you get for the matrix? Please do it yourself first before looking at the spoiler text below :-).

One can rewrite the rotational energy as follows:
$$ E = \frac12 \vec \omega^\mathrm T \mathbf I \vec \omega = \frac12 (\mathbf A \mathbf A^{-1} \vec \omega)^\mathrm T  \mathbf I \mathbf A \mathbf A^{-1} \vec \omega = \frac12 \underbrace{\vec \omega^\mathrm T \mathbf A^{-1\,\mathrm T}}_{\vec \omega_1^\mathrm T} \underbrace{\mathbf A^\mathrm T \mathbf I \mathbf A}_{\mathbf I_1} \underbrace{\mathbf A^{-1} \vec \omega}_{\vec \omega_1} \,. $$
So one would rather say that $I_1 = \mathbf A^\mathrm T \mathbf I \mathbf A$ in the general case. However, a transformation which you use in cases like these are orthogonal transformations. Those are just rotations, so they are a physical symmetry of the system. Also the matrix $\mathbf I$ is symmetric, therefore it can be diagonalized with an orthogonal transformation. Therefore we may assume that $\mathbf A \in \mathrm{SO}(3)$ and therefore $\mathbf A^\mathrm T$ = $\mathbf A^{-1}$. Then we do get the solution that I wrote previously: $\mathbf I_1 = \mathbf A^{-1} \mathbf I \mathbf A $.
Going the other way will give $\mathbf I = \mathbf A \mathbf I_1 \mathbf A^{-1} $.

This differs from Rod's answer a bit because he directly obtains $\mathbf I_1 = \mathbf A^{-1} \mathbf I \mathbf A$ without having to assume $\mathbf A \in \mathrm{SO}(3)$. The reason for this is really subtle and will only become clear once you learn about special relativity with four-vectors and the metric tensor.
I have just assumed that the rotational energy will stay the same in every coordinate system. That is not always the case, that is only the case for transformations which leave the scalar product invariant. This is somewhat of a tautology, therefore I missed in the first go.
The invariance of scalar products depends on the invariance of the metric tensor. Only orthogonal transformations, those from SO(3), leave this metric tensor invariant. So in one has to restrict coordinate transformations right from the beginning to those.
Once you looked into special relativity, you will get that the group of symmetry operations for spacetime is something like $\mathrm{SO}(1, 3) \times \mathbb R^4$. So one directly restricts to special orthogonal matrices from the start, the scalar product is left invariant and one can do the argumentation that I did earlier.
Rod needs to assume way less in order to derive the result. So I think his answer is the cleaner one as I have made too strong assumptions.
A: For what it's worth, here's how you work it out from the transformation of the angular momentum - I began this answer last night - but here is how Martin's Answer would play out with deducing the transformation laws from the assumption of vector transformation laws for the angular momentum and angular velocity instead of the assumption of energy's being a scalar and angular velocity a vector as in Martin's answer.
The definition of the inertia tensor is the linear homogeneous transformation characterizing a body that maps the angular velocity vector to the angular momentum:
$$\vec{L} = I\,\vec{\omega}\tag{1}$$
Vectors, by your definition of symbols, in this picture transform as $X \mapsto A\,X$. So we write $\vec{L} = A\,\vec{L}^\prime$, likewise $\vec{\omega} = A\,\vec{\omega}^\prime$, where the primed vectors are the components in the new co-ordinates. Substitution into (1) and fore-multiplying both sides by $A^{-1}$ (naturally your transformation is nonsingular) yields:
$$\vec{L}^\prime = A^{-1}\,I\,A\,\vec{\omega}^\prime$$
whence you can read off the transformed inertia matrix from the definition given in (1) and naturally this is the same as Martin's answer.
