Normal subgroup and non-equivalent representations I am looking for an example to illustrate that to representations are not equivalent in a particular situation.
If H is a normal subgroup of G, $\pi$ a representation of $H$, and the second representation is defined as follows: For $g \in G$ define $\pi^g : G \to GL(V)$ by $\pi^g (h)=\pi (g^{-1}hg)$. For $g \in G$ we define the representation $\pi^g$ of H as before.
So now I have to find an example of G, H a representation $\pi$ of H and $g \in G$ such that $\pi$ and $\pi^g$ are not equivalent.
However, I am struggling a bit to find a nice and simple example to show this.
I've tried with $G= (\mathbb{C}, +)$ and $H=(\mathbb{R}, +)$ as this is a normal subgroup of G (as it is abelian) and as the representation of H has been given in the lectures. However, it seems as if this doesn't work
 A: Since $\pi(g^{-1}hg) = \pi(g)^{-1}\pi(h)\pi(g)$, you're just conjugating the outputs, i.e. you're using $\pi(g)$ to change bases. So, they're equivalent representations.

Edit: As the comments noted, you intended to write $\pi^g \colon H \to GL(V)$, in which case they don't need to be equivalent.
For instance, take $G = S_3$, the symmetric group on 3 elements, and $H = A_3 = \{1, (123), (132)\}$, the alternating group on 3 elements, written in cycle notation. Use $V = \mathbb{C}$ and $\pi \colon A_3 \to \mathbb{C}^\times$ by $\pi((123)^k) = \exp(2\pi i k/3)$.
Set $g = (12)$. Easy calculations give $\pi^g((123)) = \pi((123)^2) = \exp(2\pi i \cdot 2/3) = \overline{\pi((123))}$, the complex conjugate. Hence they're inequivalent representations.
A: Let $V=\mathbb{V}_2(\mathbb{F}_4)$, let $H:=\text{GL}(V)$ be the general linear group, and let $G:=\text{$\Gamma$L}(V)$ be the semi-linear group, that is $H$ extended by the Frobenius automorphism $\sigma$.
Now let $\pi:H\to \text{GL}(V)$ be given by $h\mapsto h$. The representation $\pi^{\sigma}$ will not be equivalent to $\pi$.
