Let $H$ be a subgroup of a finite group $G.$ Suppose $(\sigma, W)$ and $(\sigma', W')$ be representation of $H$ over $\mathbb{C}$. Let $H$ be a subgroup of a finite group $G$, and $(\sigma,W)$ and $(\sigma'W')$ be representation of $H$ over $\mathbb{C}$.
Let ${\rm Ind}_H^G \sigma$ be  $${\rm Ind}_H^G \sigma = \{f:G \rightarrow W | f(hg) = σ(h) f(g) \ \  \forall h\in H, g \in G \}$$
Then, if $\sigma$ and $\sigma'$ are isomorphism of representation, why ${\rm Ind}_H^G \sigma$ and ${\rm Ind}_H^G \sigma'$ is isomorphic ?
I want to construct concrete isomorphism between them.
Thank you in advance.
 A: Probably the only reason you are confused is because here $\sigma$ and $\sigma'$ have the same underlying vector space $W$. For clarity, let's have $\phi : W \to W'$ be an isomorphism of the representations $(\sigma, W)$ and $(\sigma', W')$. Then elements of $\operatorname{Ind}^G_H \sigma$ are maps $f : G \to W$, while elements of $\operatorname{Ind}^G_H \sigma'$ are maps $g : G \to W'$. An isomorphism of these induced representations must turn maps of the former kind into maps of the latter kind, so given the data we have available (only $\phi$) how can we do this?
Spoiler: The only way I can think of is to map $f \mapsto \phi \circ f$. We need to check that actually $\phi \circ f \in \operatorname{Ind}^G_H \sigma'$, but this is essentially by definition. Then since $\phi$ is invertible, the assignment $f \mapsto \phi \circ f$ is invertible too. It just remains to prove that the assignment $f \mapsto \phi \circ f$ intertwines $\operatorname{Ind}^G_H \sigma$ and $\operatorname{Ind}^G_H \sigma'$, but there isn't really anything to do here either.
