Yes. Let $$ F(g) = \sum_{h \in G} \overline{\chi(h)}\chi(gh).$$ I claim that $F$ is a class function. Indeed, $$F(s^{-1} g s) = \sum_{h \in G} \overline{\chi(h)}\chi(s^{-1}gsh)=\sum_{h \in G} \overline{\chi(h)}\chi(g(shs^{-1})) = \sum_{h' \in G} \overline{\chi(s^{-1}h's )}\chi(gh') = \sum_{h' \in G} \overline{\chi(h' )}\chi(gh') = F(g).$$

Thus, we can write $F$ as a combination of irreducible characters: $$F(g) = \sum_{i} a_{\chi_i} \chi_i(g).$$ Let $\rho_{\chi_i}$ be the representation with character $\chi_i$. Then $$a_{\chi_i} = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}F(g) = \frac{1}{|G|} \sum_{g,h} \overline{\chi_i(g)} \overline{\chi(h)}\chi(gh)= \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g) ). $$ Letting $$A_i = \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g),$$ we have $$(*)a_{\chi_i} = \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) A_i).$$ I claim $A_i$ commutes with $\rho_{\chi}(h)$ for all $h$-s: $$A_i \rho_{\chi}(h) =\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(gh) =\sum_{g'} \overline{\chi_i(g'h^{-1})}\rho_{\chi}(g')=\sum_{g'} \overline{\chi_i(h^{-1} g')}\rho_{\chi}(g')=\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(hg)= \rho_{\chi}(h)A_i.$$ Thus, by Schur's lemma, $A_i$ is a multiple of the identity: $$A_i = c_i I$$ for some constant $c_i$, which can be extracted by taking traces: $$c_i \cdot \dim(\rho_i) = \mathrm{Tr}(A_i) = \sum_{g \in G}\overline{\chi_i}(g) \chi(g) = 1_{\chi_i = \chi} |G|.$$ Thus, when $\chi_i \neq \chi$ we have $A_i=0$ and so $a_{\chi_i}=0$ in that case by $(*)$, and $F(g)$ must be proportional to $\chi(g)$. To find the constant of proportionality, note that if $\chi_i=\chi$, we have from $(*)$ that $$a_{\chi_i}=\frac{1}{\dim \rho_{\chi_i}}\sum_{h} \overline{\chi_i}(h) \chi(h) = \frac{|G|}{\chi(e)},$$ as needed.

Yes. Let $$ F(g) = \sum_{h \in G} \overline{\chi(h)}\chi(gh).$$ I claim that $F$ is a class function. Indeed, $$F(s^{-1} g s) = \sum_{h \in G} \overline{\chi(h)}\chi(s^{-1}gsh)=\sum_{h \in G} \overline{\chi(h)}\chi(g(shs^{-1})) = \sum_{h' \in G} \overline{\chi(s^{-1}h's )}\chi(gh') = \sum_{h' \in G} \overline{\chi(h' )}\chi(gh') = F(g).$$

Thus, we can write $F$ as a combination of irreducible characters: $$F(g) = \sum_{i} a_{\chi_i} \chi_i(g).$$ Let $\rho_{\chi_i}$ be the representation with character $\chi_i$. Then $$a_{\chi_i} = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}F(g) = \frac{1}{|G|} \sum_{g,h} \overline{\chi_i(g)} \overline{\chi(h)}\chi(gh)= \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( [\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g)]\rho_{\chi}(h) ). $$ Letting $$A_i = \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g),$$ we have $$(*)a_{\chi_i} = \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( A_i\rho_{\chi}(h) ).$$ I claim $A_i$ commutes with $\rho_{\chi}(h)$ for all $h$-s: $$A_i \rho_{\chi}(h) =\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(gh) =\sum_{g'} \overline{\chi_i(g'h^{-1})}\rho_{\chi}(g')=\sum_{g'} \overline{\chi_i(h^{-1} g')}\rho_{\chi}(g')=\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(hg)= \rho_{\chi}(h)A_i.$$ Thus, by Schur's lemma, $A_i$ is a multiple of the identity: $$A_i = c_i I$$ for some constant $c_i$, which can be extracted by taking traces: $$c_i \cdot \dim(\rho_i) = \mathrm{Tr}(A_i) = \sum_{g \in G}\overline{\chi_i}(g) \chi(g) = 1_{\chi_i = \chi} |G|.$$ Thus, when $\chi_i \neq \chi$ we have $A_i=0$ and so $a_{\chi_i}=0$ in that case by $(*)$, and $F(g)$ must be proportional to $\chi(g)$. To find the constant of proportionality, note that if $\chi_i=\chi$, we have from $(*)$ that $$a_{\chi_i}=\frac{1}{\dim \rho_{\chi_i}}\sum_{h} \overline{\chi_i}(h) \chi(h) = \frac{|G|}{\chi(e)},$$ as needed.

Yes. Let $$ F(g) = \sum_{h \in G} \overline{\chi(h)}\chi(gh).$$ I claim that $F$ is a class function. Indeed, $$F(s^{-1} g s) = \sum_{h \in G} \overline{\chi(h)}\chi(s^{-1}gsh)=\sum_{h \in G} \overline{\chi(h)}\chi(g(shs^{-1})) = \sum_{h' \in G} \overline{\chi(s^{-1}h's )}\chi(gh') = \sum_{h' \in G} \overline{\chi(h' )}\chi(gh') = F(g).$$

Thus, we can write $F$ as a sum of irreducible characters: $$F(g) = \sum_{i} a_{\chi_i} \chi_i(g).$$ Let $\rho_{\chi_i}$ be the representation with character $\chi_i$. Then $$a_{\chi_i} = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}F(g) = \frac{1}{|G|} \sum_{g,h} \overline{\chi_i(g)} \overline{\chi(h)}\chi(gh)= \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g) ). $$ Letting $$A_i = \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g),$$ we have $$(*)a_{\chi_i} = \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) A_i).$$ I claim $A_i$ commutes with $\rho_{\chi}(h)$ for all $h$-s: $$A_i \rho_{\chi}(h) =\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(gh) =\sum_{g'} \overline{\chi_i(g'h^{-1})}\rho_{\chi}(g')=\sum_{g'} \overline{\chi_i(h^{-1} g')}\rho_{\chi}(g')=\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(hg)= \rho_{\chi}(h)A_i.$$ Thus, by Schur's lemma, $A_i$ is a multiple of the identity: $$A_i = c_i I$$ for some constant $c_i$, which can be extracted by taking traces: $$c_i \cdot \dim(\rho_i) = \mathrm{Tr}(A_i) = \sum_{g \in G}\overline{\chi_i}(g) \chi(g) = 1_{\chi_i = \chi} |G|.$$ Thus, wen $\chi_i \neq \chi$ we have $A_i=0$ and so $a_{\chi_i}=0$. If $\chi_i=\chi$, we have from $(*)$ $$a_{\chi_i}=\frac{1}{\dim \rho_{\chi_i}}\sum_{h} \overline{\chi_i}(h) \chi(h) = |G|/\chi(e),$$ as needed.

Yes. Let $$ F(g) = \sum_{h \in G} \overline{\chi(h)}\chi(gh).$$ I claim that $F$ is a class function. Indeed, $$F(s^{-1} g s) = \sum_{h \in G} \overline{\chi(h)}\chi(s^{-1}gsh)=\sum_{h \in G} \overline{\chi(h)}\chi(g(shs^{-1})) = \sum_{h' \in G} \overline{\chi(s^{-1}h's )}\chi(gh') = \sum_{h' \in G} \overline{\chi(h' )}\chi(gh') = F(g).$$

Thus, we can write $F$ as a combination of irreducible characters: $$F(g) = \sum_{i} a_{\chi_i} \chi_i(g).$$ Let $\rho_{\chi_i}$ be the representation with character $\chi_i$. Then $$a_{\chi_i} = \frac{1}{|G|} \sum_{g \in G} \overline{\chi_i(g)}F(g) = \frac{1}{|G|} \sum_{g,h} \overline{\chi_i(g)} \overline{\chi(h)}\chi(gh)= \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g) ). $$ Letting $$A_i = \sum_{g} \overline{\chi_i(g)}\rho_{\chi}(g),$$ we have $$(*)a_{\chi_i} = \frac{1}{|G|} \sum_{h} \overline{\chi(h)} (\mathrm{Tr}( \rho_{\chi}(h) A_i).$$ I claim $A_i$ commutes with $\rho_{\chi}(h)$ for all $h$-s: $$A_i \rho_{\chi}(h) =\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(gh) =\sum_{g'} \overline{\chi_i(g'h^{-1})}\rho_{\chi}(g')=\sum_{g'} \overline{\chi_i(h^{-1} g')}\rho_{\chi}(g')=\sum_{g} \overline{\chi_i(g)}\rho_{\chi}(hg)= \rho_{\chi}(h)A_i.$$ Thus, by Schur's lemma, $A_i$ is a multiple of the identity: $$A_i = c_i I$$ for some constant $c_i$, which can be extracted by taking traces: $$c_i \cdot \dim(\rho_i) = \mathrm{Tr}(A_i) = \sum_{g \in G}\overline{\chi_i}(g) \chi(g) = 1_{\chi_i = \chi} |G|.$$ Thus, when $\chi_i \neq \chi$ we have $A_i=0$ and so $a_{\chi_i}=0$ in that case by $(*)$, and $F(g)$ must be proportional to $\chi(g)$. To find the constant of proportionality, note that if $\chi_i=\chi$, we have from $(*)$ that $$a_{\chi_i}=\frac{1}{\dim \rho_{\chi_i}}\sum_{h} \overline{\chi_i}(h) \chi(h) = \frac{|G|}{\chi(e)},$$ as needed.