Character does not determine representation in any prime characteristic

Statement
Suppose $$G$$ is a finite group and $$K$$ is a field of characteristic $$p$$ a prime number (it does not matter here whether $$p$$ divides the order of $$G$$). It is possible to construct two representations $$\varphi_1,\varphi_2$$ of $$G$$ over $$K$$ that are inequivalent as representations but have the same character.

Related facts

 * Character determines representation in characteristic zero

Outrageous example
Take $$\varphi_1$$ as a sum of $$p$$ copies of the trivial representation and $$\varphi_2$$ as a sum of $$2p$$ copies of the trivial representation.

Both $$\varphi_1$$ and $$\varphi_2$$ have character value zero everywhere, but they are not equivalent -- in fact, they do not even have the same degree.

It's possible to construct examples of inequivalent representations with the same degree by using two distinct irreducible representations.