$A$ be a linear map on $\mathbb{C}^n$ with Charpoly $(x-1)^n$, $A$ is similar to $A^{-1}$ $A$ be a linear map on $\mathbb{C}^n$ with Charpoly $(x-1)^n$, I need to show $A$ is similar to $A^{-1}$
I thought showing $A$ and $A^{-1}$ has same Charpoly
$(A-I)^n=0\Rightarrow (A-AA^{-1})^n=0\Rightarrow A^n(I-A^{-1})=0\Rightarrow (A^{-1}-I)^n=0$
am i right? Thanks for helping
 A: As said in the comments, your argument is not a solution. However, it can be used to give a full answer.
We may assume that $A$ is in Jordan form. All entries on the diagonal are equal to $1$, and we only need to verify that the Jordan form of $A^{-1}$ has the same block sizes as $A$. Thus we may assume that $A$ has one Jordan block, and it is sufficient to show that so does $A^{-1}$. Your argument shows that $A-I$ and $A^{-1}-I$ have the same nilpotence degree. The proposition follows.
A: $A$ with characteristic polynomial $(x-1)^n$ means all the eigenvalues are $1$. $A$ is conjugate to a block diagonal matrix with diagonals Jordan cells corresponding to $\lambda=1$.  It's enough to prove that a Jordan matrix $J_{1,m}$ is similar to $J_{1,m}^{-1}\ $. Now $J_{1,m}= I + N$ where $N$ is the off diagonal part, a nilpotent matrix : $N^m = 0$. We get 
$$(I+N)^{-1} = I - N + N^2 - N^3 + \ldots + (-1)^{m-1}N^{m-1}$$
For example, 
$$J_{1,5} = \left(\begin{array}{ccccc}1 & 0 & 0 & 0 & 0 \\1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 \end{array}\right)
$$
with inverse
$$J_{1,5}^{-1} = \left(\begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 0 \\ -1 & 1 & -1 & 1 & 0 \\ 1 & -1 & 1 & -1 & 1 \end{array}\right)$$
To show that $J_{1,m}$ and $J_{1,m}^{-1}$ are conjugate it's enough to show that $N_m =J_{1,m}- I_m$ and $N' =J_{1,m}^{-1}-I_m$ are. 
The linear transformation $N_m =J_{1,m}-I_m$ maps the vectors in the following chain
$$e_1\mapsto e_2 \mapsto \ldots \mapsto e_m \mapsto 0$$
Therefore, consider the sequence of vectors
$$e_1, N' e_1, N'^2 e_1, \ldots N'^{m-1} e_1 $$
denoted by $f_1$, $f_2$, \ldots $f_m$. The linear transformation $N'$ maps these vectors in the following chain
$$f_1\mapsto f_2 \mapsto \ldots \mapsto f_m \mapsto 0$$
Once we know that the $f_i$ also form a basis we are done. In fact, this will work for any $N'$ that is strictly lower triangular and the first shifted diagonal has all the elements nonzero.  Consider the linear invertible map 
taking $e_i \mapsto f_i$. Then one checks right away that 
$$N'B = BN$$
by verifying the equality on basis vectors $e_i$. Therefore 
$N' = BNB^{-1}$ and so 
$$J_{1,m}^{-1} = B J_{1,m}B^{-1}$$
It's instructive to do this for say $m=5$. We get 
$$B= \left(\begin{array}{ccccc}1 & 0 & 0 & 0 & 0 \\
 0 & -1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & -1 & -2 & -1 & 0 \\ 0 & 1 & 3 & 3 & 1 \\ \end{array}\right)$$
and we have
$$B \cdot J_{1,5}= B'\colon =\left( \begin{array}{ccccc} 1 & 0 & 0 & 0 & 0 \\ -1 & -1 & 0 & 0 & 0 \\ 1 & 2 & 1 & 0 & 0 \\ -1 & -3 & -3 & -1 & 0 \\ 1 & 4 & 6 & 4 & 1 \\\end{array}\right)$$
Both $B$ and $B'$ are basically coming from Pascal's triangle and the above equality is the fundamental recurrence for binomial coefficients. Moreover, both $B$ and $B'$ are involutions as one can check readily. 
This holds for every $m$. From this we get 
$$J^{-1} = B J B^{-1} = BJB$$
In other way, $J = B B'$ a product of two involutions, so it is conjugate to its inverse $J^{-1}$ (in an explicit way).
All this works in general for all $m$. 
A general  observation about conjugacy classes. If two matrices are invertible and conjugate, their inverses are also conjugate. We thus have a mapping from conjugacy classes of matrices to conjugacy classes. Since the conjugacy classes of matrices are described by Jordan forms the problem reduces to finding the Jordan form of the inverse of a Jordan form, and reduces further to Jordan blocks. More generally, the Jordan form of $J_{\lambda, m}^{-1}$ is $J_{\lambda^{-1}, m}$, with an easy extension of the above.
One can also consider maps more general than $A\mapsto A^{-1}$, like polynomial or rational. 
