How to rotate a tensor? If I have a rotor in geometric algebra:
$$
R(\theta) = e^{-\mathbf{\Omega} \frac{\theta}{2}} 
$$
I know how to rotation a vector
$$
v'=RvR^{-1}=e^{-\mathbf{\Omega} \frac{\theta}{2}}  v e^{\mathbf{\Omega} \frac{\theta}{2}} 
$$
But what if I want to rotate a tensor (say of rank-2) instead of a vector. What is the operation?
 A: Let's start with a rank-1 covariant tensor, which can be defined as a linear map
$$f:V\to\mathbb{R}: v \mapsto f(v)$$
Now, how would one define a rotated version of this map? Well, presumably, if we rotate that tensor, and we rotate the inputs as well, the results should be the same as applying the original map to the unrotated vector. Thus, let $f':V\to\mathbb{R}$ be the rotated map, then
$$f'(v') = f(v)$$
or
$$f'(RvR^{-1})=f(v)$$
or
$$f'(v')=f(R^{-1}v'R)$$
Let's accept the following notational convention: $f'=f(R^{-1}.R)$.
Likewise, for a rank-2 covariant tensor $g:V\times V \to \mathbb{R}$, one would have that the rotated map $g'$ satisfies
$$g'(v,w) = g(R^{-1}vR,R^{-1}wR)$$
Now, let's go back to a rank-1 contravariant tensor, i.e. a vector. But the definition as a map would be the following:
$$v:V^* \to \mathbb{R}: f \mapsto f(v)$$
Reasoning as previously, the rotated vector $v'$ must be such that
$$v'(f')=v(f)$$
And indeed,
$$v'(f')=f'(v')=f(R^{-1}v'R)$$
which makes us recover $R^{-1}v'R=v$ and thus $v'=RvR^{-1}$.
Therefore, for a rank-2 contravariant tensor $w:V^*\times V^* \mapsto \mathbb{R}$, we have that
$$w'(f',g') = w(f,g)$$
thus
$$w'(f',g') = w(f'(R.R^{-1}),g'(R.R^{-1})) \; .$$
