Let a symmetric positive definite matrix $\boldsymbol{S}\in\mathbb{R}^{d\times d}$ be given.
Consider the following transformation: tanh is applied on the off-diagonal entries and the diagonal entries are set to $1$. Is the resulting matrix again positive definite?
Consider the matrix $$ S_n=\begin{pmatrix} 4n & 2n & n \\ 2n & 4n & 2n\\ n &2n &4n \end{pmatrix} $$ Then $S_n$ is strictly positive definite as the diagonal is dominating. The new matrix has the form $$T_n= \begin{pmatrix} 1 & a_n & b_n \\ a_n & 1 & a_n \\ b_n &a_n &1 \end{pmatrix} \qquad a_n={e^{2n}-e^{-2n}\over e^{2n}+e^{-2n}},\quad b_n={e^{n}-e^{-n}\over e^{n}+e^{-n}}$$ Then $$\displaylines{\det T_n= (b_n-1)\,[2a_n^2-b_n-1)\\ =(b_n-1)[(1-b_n)-2(1-a_n)(1+a_n)]}$$ We have $b_n<1$ and $$1-a_n\approx 2e^{-2n},\qquad 1-b_n\approx 2e^{-n}$$ Hence $$2a_n^2-b_n-1\approx 2e^{-n} -8e^{-2n}$$ Thus $\det T_n<0$ for large $n.$