I would like to have examples of a finite group, $G$, with a finite dimensional representation, $V$, (over the complex numbers, say) with four conditions:
You can interchange the role of symmetric and exterior squares, if you wish.
Let $G=\mathrm{PSU}(3,5)$, the simple group of order 126,000. Let $\chi_2$ be the $20$ dimensional representation. This has an invariant (symplectic) inner product. Then the symmetric square is $\chi_7 + \chi_{10}$, while the exterior square is $\chi_1+\chi_4+\chi_5+\chi_6+\chi_8$. Here, the dimensions are $84+126$ and $1+28+28+28+105$. This is an example with the roles interchanged, as permitted by the question. I've used ATLAS notation for the characters.
Here is a simple infinite family of examples which shows that we can have as many irreducible constituents as we want in both the symmetric and exterior square. Let $G=\operatorname{PGL}_2(\mathbb{F}_q)$, where $q$ is odd and let $V$ be the Steinberg representation (which has dim $q$). It can be checked that $V\otimes V$ is multiplicity-free and contains every irrep of $G$ except the non-trivial one-dimensional one (this is the least obvious step). Moreover, $V$ has real-valued character, so carries a $G$-invariant bilinear form.
The group $G$ has $(q-3)/2$ irreps of dim $q+1$, $(q-1)/2$ irreps of dim $q-1$, 2 irreps of dim $q$ and 2 of dim 1. Meanwhile, the symmetric square of $V$ has dim $q(q+1)/2$ and the exterior square has dim $q(q-1)/2$. Thus, for $q$ large enough, each one contains more than $N$ irreducible constituents, for any given positive $N$. For example, for $q=11$, $G$ has order 1,320 and the symmetric and exterior squares of $V$ both have at least 5 irreducible constituents (one can easily find a better bound or even the exact numbers).
