Let $V$ be a three-dimensional Euclidean space. Let $T$ be a congruent transformation over $V$. Let $v$ and $w$ be two vectors in $V$. Is there an interesting relation between $(Tv)\wedge (Tw)$ and $T(v\wedge w)$?
This is an exercise from an introductory book on differential geometry.
If $v\wedge w$ is vector to you, then $T(v\wedge w) = \pm T(v)\wedge T(w)$, where you can decide which sign you get by finding the determinant of $T$.
For reference, the general way to calculate $T(v)\wedge T(w)$ is by finding the matrix $M$ which represents $T$, then the formula for any linear transformation $T$ is given by $$(Mv)\wedge (Mw)=\det(M)M^{-T}(v\wedge w)$$
If $v\wedge w$ is a bivector to you, then technically the same transformation $T$ can't act on both vectors and bivectors. If you see $T(v\wedge w)$, what this actually means is the outermorphism of the linear transformation $T$. When that's clear, often the same notation is used for both the linear transformation and its outermorphism. When it's not, we use some other notation: I prefer $\underline T$. The outermorphism $\underline T$ is defined as the unique linear transformation satisfying:
$$ \underline{\mathsf{T}}(x) = T(x)\\ \underline{\mathsf{T}}(A \wedge B) = \underline{\mathsf{T}}(A) \wedge \underline{\mathsf{T}}(B)\\ \underline{\mathsf{T}}(A + B) = \underline{\mathsf{T}}(A) + \underline{\mathsf{T}}(B)\\ \underline{\mathsf{T}}(1) = 1 $$
for any multivectors $A$ and $B$ and any vector $x$. Thus, by definition $\underline T(v\wedge w) = T(v)\wedge T(w)$ for any type of linear transformation -- including congruent transformations.