Nonisomorphic representations become isomorphic when induced? 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
$$L(s,\pi) = L(s,\operatorname{Ind}_H^G \pi) = L(s,\operatorname{Ind}_H^G \pi \circ i_{\sigma}) = L(s, \pi \circ i_{\sigma})$$
 A: 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 $\operatorname{Ind}_H^G \pi \cong \operatorname{Ind}_H^G (\pi \circ i_s)$.
A: 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.
