Let $A$ be a complex Hermitian $n\times n$ matrix and define the matrix $B$ to be the entry-wise absolute value of $A$, i.e., $B_{ab}=\lvert A_{ab}\rvert$. Furthermore suppose that $B$ has a unique normalised eigenvector $x$ of maximal eigenvalue $\lambda>0$, $Bx=\lambda x$, in particular $\lVert B\lVert=\lambda$ (here $\lVert \cdot\rVert$ denotes the induced matrix norm from the Euclidean norm on $\mathbb C^n$).
It is obvious that $\lVert A\rVert\le \lVert B\rVert$. I now first want to consider the extreme case when $\lVert A\rVert=\lVert B\lVert$. It follows that $A$ has an normalized eigenvector $y$ of eigenvalue $\lambda$, $Ay=\lambda y$ and we can compute \begin{align*} \lambda =\langle y, Ay\rangle= \sum_{ab}\overline{y_a} A_{ab}y_b\le \sum_{ab} \lvert y_a\rvert B_{ab} \lvert y_b\rvert =\langle\lvert y\rvert,B \lvert y\rvert\rangle\le \lambda =\langle x,Bx\rangle.\end{align*} In particular, it follows that $\lvert y\rvert =x$.
Now I am interested in how this can be made quantitative. For example, if $\lVert A\rVert \ge (1-\epsilon)\lVert B\rVert$, is it true that $\lVert \lvert y\rvert -x\rVert\le C \epsilon$ for some universal constant $C$? I think the above argument can be made quantitative to give $\lVert \lvert y\rvert -x\rVert\le C \sqrt\epsilon$, where $C$ depends on the spectral gap of $B$ below the eigenvalue $\lambda$.