Nonisomorphic representations become isomorphic when induced?

Question

Let $H$ be a subgroup of a group $G$, and let $s \in G$. Assume that $sHs^{-1} = H$. Let $i_s$ be the automorphism of $G$ given by $x \mapsto sxs^{-1}$. Then this restricts to an automorphism of $H$.

Let $(\pi,W)$ be a representation of $H$. There is no reason that the representations $\pi$ and $\pi \circ i_s$ should be isomorphic. But do we have $\operatorname{Ind}_H^G \pi \cong \operatorname{Ind}_H^G (\pi \circ i_s)$?

Motivation: let $E$ be a Galois extension of local field $F$, $H = \operatorname{Gal}(\overline{F}/E), G = \operatorname{Gal}(\overline{F}/F)$, $\sigma \in G$, and $\pi$ a continuous finite dimensional representation of $H$. I want to say that the L-functions $L(s,\pi)$ and $L(s,\pi \circ i_{\sigma})$ are equal. Since L-functions are inductive, if we assume the given result is true, then we have

$$L(s,\pi) = L(s,\operatorname{Ind}_H^G \pi) = L(s,\operatorname{Ind}_H^G \pi \circ i_{\sigma}) = L(s, \pi \circ i_{\sigma})$$

Answer 1

For a very simple example, take $G=A_4$ and $H=V_4$, its normal, elementary Abelian, Sylow $2$-subgroup. Then $H$ has three non-trivial degree one representations, each with kernel one of the three order $2$ subgroups of $V_4$. Under conjugation by an order $3$ element of $A_4$ but as representations of $V_4$ they are non-isomorphic. However, when induced to $A_4$, each of them yields the same representation, up to isomorphism.

Answer 2

Yes. Let $V_1, V_2$ be the respective underlying spaces of $\operatorname{Ind}_H^G \pi$ and $\operatorname{Ind}_H^G (\pi \circ i_s)$. Then $V_1$ (resp. $V_2$) consists of all functions $f: G \rightarrow W$ such that $f(hg) = \pi(h)f(g)$ (resp. $f(hg) = \pi(shs^{-1})f(g)$) for all $h \in H, g \in G$. The group $G$ acts on each of these spaces by right translation.

Define a linear isomorphism $\Phi: V_1 \rightarrow V_2$ by $\Phi(f)(x) = f(sx)$. Let's check that if $f \in V_1$, $\Phi(f)$ is actually in $V_2$:

$$\Phi(f)(hg) = f(shg) = f(shs^{-1}sg) = \pi(shs^{-1})f(sg) = \pi(shs^{-1})\Phi(f)(g)$$

The inverse $\Phi^{-1}:V_2 \rightarrow V_1$ is given by $\Phi^{-1}(f)(x) = f(s^{-1}x)$.

Let's check that $\Phi$ intertwines the action of $G$. Let $f \in V_1, g \in G$. For all $x \in G$, we have

$$\Phi(g \cdot f)(x) = g \cdot f(sx) = f(sxg)$$

$$g \cdot \Phi(f)(x) = \Phi(f)(xg) = f(sxg)$$

and therefore $\Phi(g \cdot f) = g \cdot \Phi(f)$. This shows that $\operatorname{Ind}_H^G \pi \cong \operatorname{Ind}_H^G (\pi \circ i_s)$.