Matrix similarity by changing basis 
Let $$A= \pmatrix{0&1&1\\1&0&0\\2&1&0}$$ and $$B = \pmatrix{0&1&0\\1&0&1\\1&2&0}.$$ Show that $A$ and $B$ are similar in $\mathbb R$.

We can do this by showing that $A$ and $B$ are similar to the same diagonal matrix: they have the same characteristic polynomial, i.e. $$\chi = X³ - 3X - 1.$$
We can then painfully show that $\chi$ has three distincs real roots $\alpha, \beta, \gamma$ so $A$ and $B$ are similar to $$D = \pmatrix{\alpha&0&0\\0&\beta&0\\0&0&\gamma}.$$
Hence they are similar.
I want to show that $A$ and $B$ are similar using a simpler method:
Let $u$ be the linear map such that $A$ is the matrix of $u$ in basis $\mathcal B = (e_1, e_2, e_3)$.
My question is how can I find a basis $\mathcal C = (f_1, f_2, f_3)$ such that $u$ has matrix $B$ in this basis?
 A: Observe that one matrix is obtained from the other by swapping the first two columns and then the first two rows.
This means that the permutation matrix $P=\pmatrix{0&1&0\\1&0&0\\0&0&1}$ is the change of basis that you are after, in other words: $$A=PBP^{-1}$$ 
where moreover $P^{-1}=P$. Multiplication by $P$ on the right will swap the columns of $B$, and then multiplication on the left will swap the rows. 
See another similar (!) question here.
A: Note that these matrices contain the same entries. Thus, as a first guess, there might be a permutation given you the similarity (yes, there is, I checked^^).
In general, checking matrices for similarity is a difficult problem.
Over algebraically closed fields, Jordan's normal form is a strong tool for that, and there are also other normal forms (Frobenius, Weierstraß, etc.) that might work over other fields. But still, Jordan's normal form needs a computation of Eigenvalues, so it is not easier than what you have here.
Especially for real matrices it is almost impossible to give a practical algorithm that always solves the problem, as you can't do exact computation with real numbers.
A: You can see that
$$
u(f_1)  = u(e_1 - e_2 + e_3)\\
u(f_2) = u(2e_2 - e_3)\\
u(f_3) = u(e_1 - 2e_2 + 2e_3)
$$
therefore with
$$
f_1  = e_1 - e_2 + e_3\\
f_2 = 2e_2 - e_3\\
f_3 = e_1 - 2e_2 + 2e_3
$$
we have
$$
\text{mat}_{\mathcal C}(u) = B
$$
and
$$
\text{mat}_{\mathcal B}(C) = 
\begin{pmatrix} 
1 & 0 & 1\\
-1 & 2 & -2\\
1 & -1 & 2
\end{pmatrix}
$$
