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?

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.
