Let $f(X):\mathbb{S}^{n\times n}_{++}\rightarrow \mathbb{R}$ and $g(X):\mathbb{S}^{n\times n}_{++}\rightarrow \mathbb{R}$ be two functions where $\mathbb{S}^{n\times n}_{++}$ means the positive definite cone. If $f(X)=g(X)$ for all $X\in \mathbb{S}^{n\times n}_{++}$, does $\nabla f(X)=\nabla g(X)$?
For a more specific problem, let $\mathbf{x},\mathbf{y}\in\mathbb{R}^{m}$ and $\boldsymbol{\Sigma}\in\mathbb{S}^{2m\times 2m}_{++}$. $\boldsymbol{\Sigma}$ can be formulated as$$\boldsymbol{\Sigma}=\left[\begin{array}{cc} \mathbf{A} & -\mathbf{B}\\ \mathbf{B} & \mathbf{A} \end{array}\right],$$ where $\mathbf{A},\mathbf{B}\in\mathbb{R}^{m\times m}$, $\mathbf{A}\in\mathbb{S}^{m}$ and $\mathbf{B}$ is a skew symmetric real matrix. Let$$f\left(\boldsymbol{\Sigma}\right)=\left[\begin{array}{cc} \mathbf{x}^T & \mathbf{y}^T \end{array}\right]\boldsymbol{\Sigma}^{-1}\left[\begin{array}{c} \mathbf{x}\\ \mathbf{y} \end{array}\right]$$and$$g\left(\boldsymbol{\Sigma}\right)=\left[\begin{array}{cc} -\mathbf{y}^T & \mathbf{x}^T \end{array}\right]\boldsymbol{\Sigma}^{-1}\left[\begin{array}{c} -\mathbf{y}\\ \mathbf{x} \end{array}\right].$$ We can easily figure out that $f\left(\boldsymbol{\Sigma}\right)=g\left(\boldsymbol{\Sigma}\right)$. However, since $$\nabla f\left(\boldsymbol{\Sigma}\right)=-\boldsymbol{\Sigma}^{-1}\left[\begin{array}{c} \mathbf{x}\\ \mathbf{y} \end{array}\right]\left[\begin{array}{cc} \mathbf{x}^T & \mathbf{y}^T \end{array}\right]\boldsymbol{\Sigma}^{-1}$$and$$\nabla g\left(\boldsymbol{\Sigma}\right)=-\boldsymbol{\Sigma}^{-1}\left[\begin{array}{c} -\mathbf{y}\\ \mathbf{x} \end{array}\right]\left[\begin{array}{cc} -\mathbf{y}^T & \mathbf{x}^T \end{array}\right]\boldsymbol{\Sigma}^{-1},$$ we may have $\nabla f\left(\boldsymbol{\Sigma}\right)\neq\nabla g\left(\boldsymbol{\Sigma}\right)$. Why does this happen?