If $\pi$ is an irreducible representation of a group G such that the trivial representation occurs in $\pi \otimes \pi$ then does $\pi$ have to be self-dual, i.e $\pi \cong \pi^{*}$.$\\$
My attempt: We have $\langle \chi,\chi \rangle$=1 and also $\langle \chi_{\pi \otimes \pi},\mathbb{1} \rangle$=m (where m$\geq$1), from this to information we get: $$\frac{1}{|G|} \sum_{g\in G} |\chi(g)|^{2}=1,\frac{1}{|G|} \sum_{g\in G}( \chi(g))^{2}=m $$
If $\pi$ is self dual then $\chi_{\pi}(g)$=$\bar{\chi_{\pi}(g)}$ and then m must equals 1. So if we can get a $\pi$ such that trivial representation occurs more that once in $\pi \otimes \pi$ we are done and we can conclude that it is not self dual necessarily. But I can't find such an example and I don't know whether really we can find such an example or not. Thanks in advance!!!!
You're on a good way. So on the one hand $\langle\chi_{\pi\otimes\pi},1\rangle$ is the multiplicity of the trivial representation in $\pi\otimes\pi$. On the other hand, however, you also have $\chi_{\pi\otimes\pi} = \chi_{\pi}\chi_{\pi}$, and so $\langle\chi_{\pi\otimes\pi},1\rangle=\langle\chi_{\pi}\chi_{\pi},1\rangle=\ldots$ Can you continue from here leading you to an expression telling you whether $\pi$ and $\pi^{\ast}$ are isomorphic (remember $\langle\chi,\tau\psi\rangle=\langle\chi\overline{\tau},\psi\rangle$)?
Keeping in mind that the scalar product on characters is just a way of computing dimensions of homomorphism spaces, the above is a direct translation of the following representation theoretic content: You have $\pi\otimes\pi\cong\pi\otimes\pi^{\ast\ast}\cong\text{Hom}(\pi,\pi^{\ast})$. Now the invariants of the left hand side describe the trivial subrepresentations of $\pi\otimes\pi$, while the invariants on the right hand side are $\text{Hom}_G(\pi,\pi^{\ast})$.