Schur decomposition of a matrix with distinct eigenvalues is almost unique 
Let $M\in \mathbb C^{n,n}$ have $n$ distinct eigenvalues, and let $U_1, U_2$ be two Schur-forms of $M$. Show that if $U_1, U_2$ have equal diagonals, there is a hermitian diagonal matrix $Q$ such that $U_1=QU_2Q^H.$

My attempt: we have $M=Q_1U_1Q_1^H$ and $M=Q_2U_2Q_2^H$ for some hermitian $Q_1, Q_2$. So $$U_1 = \underbrace{Q_1^HQ_2}_{=:Q}U_2Q_2^HQ_1.$$
$Q$ is clearly hermitian but how do we show that it's diagonal? I was hoping to show that it's upper-triangular and then it would follow. Yet how do we use the setup? $U_1, U_2$ have equal diagonals, so $U_1 = D+N_1, U_2=D+N_2$ (for a diagonal $D$ and nilpotent $N_1, N_2$). From the distinctness of the eigenvalues follows that $M$ is diagonalizable but I don't quite see how to use it either. 
Any hints are hugely appreciated.
 A: Let us prove that $Q$ is upper triangular. Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb{C}^n$. Suppose $a_1,\ldots,a_n$ are the diagonal entries of $U_1$ and $U_2$.

Lemma: Let $P_{n\times n}$ be upper triangular with distinct diagonal entries $a_1,\ldots,a_n$. If $v_i$ is an eigenvector of $P$ associated to $a_i$ then $v_i\in\text{span}\{e_1,\ldots,e_i\}$.
Proof: Since $P$ is upper triangular then $P(\text{span}\{e_1,\ldots,e_i\})\subset\text{span}\{e_1,\ldots,e_i\}$ for every $1\leq i\leq n$. Notice that the eigenvalues of $P|_{\text{span}\{e_1,\ldots,e_i\}}$ are $a_1,\ldots,a_i$. Thus, there is $0\neq w_i\in \text{span}\{e_1,\ldots,e_i\}$ such that $Pw_i=a_iw_i$. Finally, if $v_i$ is an eigenvector of $P$ associated to $a_i$ then $v_i=c w_i$, since the multiplicity of $a_i$ is $1$. Therefore, $v_i\in \text{span}\{e_1,\ldots,e_i\}$. $\square$

Now, let $v_1,\ldots,v_n$ be the eigenvectors of $U_2$ associated to $a_1,\ldots,a_n$, respectively. Since $U_2$ is upper triangular, by the previous lemma, $\text{span}\{v_1,\ldots,v_i\}\subset \text{span}\{e_1,\ldots,e_i\}$. 
Now, since $\{v_1,\ldots,v_i\}$ are eigenvectors associated to different eigenvalues then they are L.I. and $\text{span}\{v_1,\ldots,v_i\}= \text{span}\{e_1,\ldots,e_i\}$.
Next, $U_1=QU_2Q^{H}$ and $U_1Qv_i=QU_2Q^{H}Qv_i=QU_2v_i=a_iQv_i$. Therefore, $\{Qv_1,\ldots,Qv_n\}$ are the eigenvectors of $U_1$ associted to $a_1,\ldots,a_n$. 
By the previous lemma, $\text{span}\{Qv_1,\ldots,Qv_i\}\subset \text{span}\{e_1,\ldots,e_i\}$. 
Finally,$Q(\text{span}\{e_1,\ldots,e_i\})=Q(\text{span}\{v_1,\ldots,v_i\})=\text{span}\{Qv_1,\ldots,Qv_i\}\subset\text{span}\{e_1,\ldots,e_i\}$. This implies that $Q$ is upper triangular.
