Show that $M=\alpha I_n\iff$ no matrix in $S$ has a zero anywhere on its diagonal. 
Let $M$ be a square complex matrix, and let $S = \{XMX^{-1} | \det(X)\neq 0 \}$ i.e.  $S$ is the set of all matrices similar to $M$
  .
Show that $M=\alpha I$ for some $\alpha\neq 0$ if and only if no matrix in $S$ has a zero anywhere on its diagonal.

The only if part is trivial. No idea on the other half...
 A: The matrix $M$ has a rational canonical form $M'$, a similar matrix
which is a diagonal sum of companion matrices. One of the companion
matrices is that associated to the minimum polynomial of $M$.
The only companion matrices without a zero on the diagonal are
$1$-by-$1$ matrices, so if $M'$ has no zero on the diagonal, its minimum
polynomial is linear, and it is a scalar matrix.
A: This was inspired in Lemma 2 of https://people.eecs.berkeley.edu/~wkahan/MathH110/trace0.pdf
Observe below that there is no restriction on the field $K$.
Claim 1: If $w, v\in K^n$ satisfy $w^Tv=1$ then $w^TMv\neq 0$.
To show that, let $\{b_2, \dots, b_n\}$ a base of the orthogonal complement of $w$. Then this set is linearly independent and also is $\{v,b_2,\dots, b_n\}$, since $v$ is not orthogonal do $w$. In particular, the matrix $B=[v | b_2 | \dots | b_n]$ is invertible. Moreover, since $w^Tv=1$ and $w^Tb_i=0$, for $i=2, \dots, n$, it follows that the first row of $B^{-1}$ is $w^T$. As a consequence, the first entry in the main diagonal of $B^{-1}MB$ is $w^TMv$, and our hypothesis implies it is nonzero and the first claim is proved.
Claim 2:  The matrix $M$ is diagonal.
Write $M=(m_{ij})$ and assume $i\neq j$ such that $m_{ij}\neq 0$. Define $w=\frac{-m_{jj}}{m_{ij}}e_i+e_j$, where $e_k$ stands for the elements of the canonical basis of $K^n$. Clearly we have for any $k$ and $l$, $e_k^TMe_l=m_{k,l}$.
Then $w^{T} e_j=1$ but
$$w^TMe_j=-\frac{m_{jj}}{m_{ij}}e_i^TMe_j+e_jMe_j=0.$$
This contradicts Claim 1. Hence $M$ is diagonal.
Claim 3: $M$ is a scalar matrix.
Write $M=diag(d_1, \dots, d_n)$, for some $d_i\in K\setminus\{0\}$. If $S$ is not scalar, there exists $i\neq j$ such that $d_i\neq d_j$. Up to a conjugation, we may assume $d_1\neq d_2$. We now consider the $n\times n$ matrix $B$ below, and its inverse
$$B=\begin{pmatrix}
 1 & \frac{d_1}{d_1-d_2} & \\
 -1& \frac{d_2}{d_2-d_1} & \\
  &                      & I_{n-2} 
 \end{pmatrix}, \qquad \qquad B^{-1}= \begin{pmatrix}
 \frac{d_2}{d_2-d_1} & \frac{d_1}{d_2-d_1} & \\
 1 & 1 &  \\
 & & I_{n-2}
 \end{pmatrix},$$ where the blank spaces are filled with $0$ and $I_{n-2}$ stands for an $n-2\times n-2$ block containing the identity matrix.
Computing $B^{-1}MB$ we obtain
$$B^{-1}MB = \begin{pmatrix}
 0 & \frac{d_1d_2}{d_2-d_1} & * \\ 
 d_1-d_2 & d_1+d_2 & *\\
 * & * & * 
 \end{pmatrix}$$ which contains 0 in the main diagonal. A contradiction to the hypothesis. Hence all diagonal entries of $M$ must be equal, which means it is a nonsingular scalar matrix.
