Is there a diagonal matrix D such that DMD is SDD, where M is SPD matrix?

Question

Let $M$ be symmetric and positive definite matrix (SPD). It is known [1] that

if $M$ is SPD and in addition satisfies $M_{ij}\leq 0$, for $i\neq j$ (called M-matrix)
then there is a positive diagonal matrix $D$ such that
$DMD$ is symmetric, diagonally dominant and $M_{ij}\leq 0$ (called SDDM matrix).

Q1: Let $M$ be SPD. Is there a diagonal matrix $D^{\prime}$ with $D^{\prime}_{ii} \neq 0~~ \forall i$ such that
$D^{\prime}MD^{\prime}$ is symmetric and diagonally dominant matrix?

Q2: Let $M$ be SPD and $[M\mathbf{1}]_i\leq[\mathrm{diag}(M)\mathbf{1}]_i~~\forall i$ (a weaker condition than $M_{ij}\leq 0$).
Is there a diagonal matrix $D^{\prime}$ with $D^{\prime}_{ii} \neq 0~~ \forall i$ such that
$D^{\prime}MD^{\prime}$ is symmetric and diagonally dominant matrix?

Q3: Is there a matrix $D^\prime$ from above such that all entries $D^{\prime}_{ii} \neq 0$ belong to
a bounded range, say $[-1,1]\backslash{\{0\}}$.

[1]: Topics in Matrix Analysis http://ebooks.cambridge.org/ebook.jsf?bid=CBO9780511840371

Answer 1

Q2: Here is a counter-example:

 1.1    -1      1
-1       1.1   -1
 1      -1      1.1

Answer 2

Q1: In general, no. Consider the all-1s matrix. It is SPSD. But, there is no scaling of it by a diagonal that makes it diagonally dominant. Adding a small amount to the diagonal makes it SPD, but doesn't change anything else.