Are all faithful, irreducible representations over $\mathbb{C}^n$ of a given finite group equivalent? Suppose I have a finite group $G$ of order $r$, and two finite-dimensional irreducible representations $D^{(1)}$ and $D^{(2)}$ both over $\mathbb{C}^n$. If these two representations are faithful, then is it true that there exists an equivariant map (or intertwining map) $S\in GL(\mathbb{C}^n)$ between the two representations?
$$S\circ D^{(1)}(g) = D^{(2)}(g)\circ S$$
Does the answer change if the two representations are instead over $\mathbb{R}^n$, but still allow the map $S$ to be complex?
$$ D^{(1)}(g), D^{(2)}(g) : \mathbb{R}^n\rightarrow \mathbb{R}^n$$
$$S:\mathbb{C}^n\rightarrow\mathbb{C}^n$$
 A: This is not true, for example faithful $1$-dimensional representations of the cyclic group of order $p$.
But it fails in a big way: let
$$m(G)=\max_{n\in \mathbb{N}} \{\text{number of inequivalent irreducible representations of $G$ of degree $n$}\}.$$
There exists a function $f:\mathbb{N}\to\mathbb{N}$ such that if $m(G)=n$ then $|G|\leq f(n)$, i.e., $|G|$ is bounded in terms of $m(G)$.
In the special case of symmetric groups, all of these (not of degree $1$) are faithful, and defined over $\mathbb{Q}$.
As two representations are equivalent over $\mathbb{C}$ if and only if they have the same character, the field over which the change of basis matrix (the intertwiner) is taken is irrevelant.
A: Consider the representations $h_1$, $h_2$ of $\mathbb{Z}/2$ defined on $\mathbb{C}^2$ by $h(1)(x,y)=(-x,y)$ and $h_2(1)(x,y)=(-x,-y)$ there are faithfull but not equivalent because the fixed of the second is $1$-dimensional and the fixed set of the first is $\{0\}$.
Consider the representation of $\mathbb{Z}/4$ defined by $h_1(1)(z)=-z$ and $h_2(1)(z)=e^{i{2\pi\over 4}}z$.
