How to find a representation with given order? This term I have to study "Representation theory of finite group", and my professor chooses the book "Representations and characters of groups" by Gordon James and Martin Liebeck, which was published by Cambridge. 
It's here :
Reprensentations and characters of groups
In this book, there are quite alot of non-trivial examples. One of them is my question. 
Firstly, from the dihedral group $D_{8}$, they give two matrices:
$A= \left[ \begin{array}{cc} 0 & 1\\ -1 & 0 \end{array} \right]$ and
$B= \left[ \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right]$ and from these two matrices, they pointed out the representation of degree 2 of $D_8$. 
And also for the representation of order 3 : $A= \left[ \begin{array}{cc} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right]$ ; $B= \left[ \begin{array}{cc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array} \right]$.
So, are there any tricks to find the matrices above ? How can we look at a finite group and find out the given order representation ?
My second question is, given two representations, how to find the invertible matrix to show that two given representations are equivalent ? It's like example $3.4$ in the book I mentioned above, this is very unnatural.
Please help me to answer these question. If I made mistakes, please feel freely helping me edit this post.
Thanks. 
 A: This is most likely what the user Sven was going to mention. It doesn't fully answer your question, but it's too long for a comment and I thought it might be useful.
So, let's go on to your first question. It is often difficult to find all the representations of a given finite group (at least practically). Often times one is presented with a group for which there are naturally equipped representations. For example, $S_n$ has the canonical one-dimensional sign representation $\text{sgn}:S_n\to\mathbb{C}^\times$, $Q_8$ (the quaternion group) has the natural matrix representation, etc. A common technique then for creating new irreps out of these canonically equipped ones is by tensoring them with one-dimensional representations. For example, there is the natural representation of $S_4$ by permuting the symbols $\{x_1,x_2,x_3,x_4\}$ of the free vector space $V=\mathbb{C}[x_1,x_2,x_3,x_4]$ and then passing to a three dimensional representation by looking at the invariant subspace of all elements of $V$ whose coefficients sum to zero (i.e. this is the standard representation). This restriction gives one a three dimensional irrep $\rho$, one can then obtain a different three-dimensional irrep by considering the tensor product $\text{sgn}\otimes\rho$. We know this is going to be irreducible. How? Because $\mathbb{C}$-representations are nice, and it suffices to check that $\chi_{\text{sgn}\otimes\rho}$ (the character associated to $\text{sgn}\otimes\rho$) is a unit vector in $\mathbb{C}[S_4]$. But, this is easy to check since $\chi_{\text{sgn}\otimes\rho}=\chi_\text{sgn}\otimes\chi_\rho$ and $\chi_\text{sgn}$ is degree one.
So, all in all, there is no fixed way to create all the representations of a given group (this should be morally true since the representation theory of a group gives a lot of the group's structure away, and so if finding the representation theory of a group is simple, so should finding the structure of the group--and this is definitely not so). 
Lastly, to address your other question. I presume what you mean is, if you suspect that two representations $\rho:G\to \text{GL}(V)$ and $\psi:G\to\text{GL}(W)$ are unitarily equivalent how does one go about finding the actual transformation $V\to W$? Once again, I don't know any hard-set rule for this (although, I am less confident that there is no 'algorithm' for this--perhaps another user has a method). That said, the beautiful thing about $\mathbb{C}$-representations is that you don't really care what the actual transformation $V\to W$ is. In fact, for all intents and purposes it is not the actual representations $\rho,\psi$ which are important, but their characters, and being equivalent unitarily (or not unitarily) implies they have the exact same characters.
I hope this helps.
