If $X^n$ is a diagonal matrix with distinct eigenvalues, then is $X$ also a diagonal matrix with distinct eigenvalues? Assume that there exists an invertible matrix $P$ such that $P^{-1}X^nP$ is a diagonal matrix with distinct eigenvalues, then can I say that $P^{-1}XP$ is also a diagonal matrix with distinct eigenvalues? If so, how do I prove it?
 A: consider nth roots of unity - $X^n=I$ but $X \ne I$. Example : $$\pmatrix{0 & 1 \\ 1 & 0 \\}^2 = \pmatrix{1 & 0 \\ 0 & 1}$$
A: Assume that $M^n$ is the diagonal matrix with diagonal $(a_j)$ of distinct entries, in particular $\prod\limits_{j}(M^n-a_j)=0$. Thus $\prod\limits_j\prod\limits_{\ell=1}^n(M-b_{j,\ell})=0$ where, for each $j$, $x^n-a_j=\prod\limits_{\ell=1}^n(x-b_{j,\ell})$, that is, the $b_{j,\ell}$ are the $n$th roots of $a_j$. Since all the coefficients $(b_{j,\ell})_{j,\ell}$ are distinct, the polynomial $\prod\limits_j\prod\limits_{\ell=1}^n(x-b_{j,\ell})$ has simple roots and is null when evaluated at $M$. Thus $M$ is diagonalizable with a diagonal of distinct entries, say, $M=QDQ^{-1}$. Hence $M^n=QD^nQ^{-1}$ and $D^n$ must have diagonal entries $(a_j)$ since $M^n$ and $D^n$ are similar. In other words, $M^n=RD^nR^{-1}$ where $R$ is a permutation matrix and $(R^{-1}Q)M^n(Q^{-1}R)=M^n$. Thus, $Q=R$, that is, $Q$ was a permutation matrix from the onset and $QDQ^{-1}$ is also a diagonal matrix.
Finally, if $M^n$ is diagonal with distinct elements, so is $M$. Using $M=A$, this answers the question the OP asked in a comment. To answer the one the OP asked in the main post, apply this to $M=P^{-1}XP$.
A: It is true that if $X^n$ is an invertible matrix, then $X$ is an invertible matrix, and that if $X^n$ has all eigenvalues distinct, then so did $X$.
What remains to show is that if $P$ is an invertible matrix that diagonalizes $X^n$ (by similarity transformation $P^{-1} X^n P$, then $P$ will also diagonalize $X$.  
If $X$ has all eigenvalues distinct, then it has a basis of eigenvectors, and we may take them as columns forming invertible matrix $Q$ such that $Q^{-1} X Q$ is diagonal.  Also $Q^{-1} X^n Q$ is diagonal, and up to a permutation of columns in $Q$, we will get the same diagonal matrix as $P^{-1} X^n P$.
Lets call that diagonal matrix $D$, bearing in mind all its diagonal entries are distinct.  Then $P^{-1} X^n P = D = Q^{-1} X^n Q$ implies $P D P^{-1} = Q D Q^{-1}$.  Hence $Q^{-1} P D = D Q^{-1} P$, and we have $D$ commutes with $Q^{-1} P$.
Claim: $R = Q^{-1} P$ is an invertible diagonal matrix.  Proof: Invertibility is clear as both $P$ and $Q$ are invertible.  It remains to show $R$ is diagonal.  Let $R_{ij}$ be an off-diagonal entry of $R$, $i \neq j$.  Then the equality $R D = D R$ means the *ij*th entry is the same, i.e. $R_{ij} D_{jj} = D_{ii} R_{ij}$.  Since those diagonal entries of $D$ are distinct, the off-diagonal entries of $R$ must be zero.  QED
Thus $P$ is $Q$ times an invertible diagonal matrix $R$, and it follows that $P$ diagonalizes $X$ just as well as $Q$ did: $P^{-1} X P = R^{-1} Q^{-1} X Q R$ is diagonal.
A: Since your question title and question body don't quite match, I will try to answer both. Presumably you are working over an algebraically closed field $\mathbb{F}$, otherwise that $X^n$ have eigenvalues in $\mathbb{F}$ doesn't necessarily imply that $X$ has eigenvalues in $\mathbb{F}$.
Now, suppose $X^n$ is a diagonal matrix that, as the question title suggests, has $n$ distinct eigenvalues. Then $X$ also has $n$ distinct eigenvalues. Therefore there exists an invertible matrix $S$ (that contains eigenvectors of $X$ as columns) and a diagonal matrix $\Lambda$ such that $X=S\Lambda S^{-1}$. By assumption, $X^n=S\Lambda^nS^{-1}$ is also a diagonal matrix. As both $X^n$ and $\Lambda^n$ are diagonal matrices sharing the same set of eigenvalues, we must have $X^n=P\Lambda^nP^T$ for some permutation matrix $P$. Therefore $S\Lambda^nS^{-1}=P\Lambda^nP^T$, or $P^TS=(\Lambda^+)^n(P^TS)\Lambda^n$, where $\Lambda^+$ denotes the Moore-Penrose pseudoinverse of $\Lambda$. Since $\Lambda$ has distinct diagonal entries, it follows that all off-diagonal entries of $P^TS$ are zero. Thus $S=PD$ for some invertible diagonal matrix $D$. Hence $X=S\Lambda S^{-1}=PD\Lambda D^{-1}P^T=P\Lambda P^T$ is a diagonal matrix. So the assertion in the question title is correct.
We now turn to the question in your question body. If $P^{-1}X^nP$ is a diagonal matrix with distinct eigenvalues, let $Y=P^{-1}XP$. Then $Y^n$ is a diagonal matrix with distinct eigenvalues. By the previous argument, $Y$ is also a diagonal matrix with distinct eigenvalues.
