Let $\Gamma_{m\times m}$ be a diagonal matrix with positive diagonal entries and $\mathbf{A}_{m\times m}$ be an arbitrary matrix. Then, is the following trace function jointly convex on $\Gamma_{m\times m}$ and $\mathbf{A}_{m\times m}$? $$\mathrm{Tr}[\mathbf{A}\Gamma^{-2}\mathbf{A}^{\mathrm{H}}\Gamma^{2}],$$ where superscript $(.)^{\mathrm{H}}$ stands for hermitian operator.

This is false.

First, $\mathrm{Tr}(AA^{\mathrm{H}}) = \|A\|_F^2$ is the square of the Frobenius norm. Then the trace has this nice circular property $\mathrm{Tr}(ABC)= \mathrm{Tr}(CAB)$, by which

$$\mathrm{Tr}[\mathbf{A}\Gamma^{-2}\mathbf{A}^{\mathrm{H}}\Gamma^{2}] = \mathrm{Tr}[\Gamma\mathbf{A}\Gamma^{-1}(\Gamma\mathbf{A}\Gamma^{-1})^{\mathrm{H}}] = \|\Gamma A\Gamma^{-1}\|_F^2.$$

Now to conclude you would take $A,B \in \mathbb{C}^{n\times n}$ and $\Gamma,\Phi$ diagonal, Hermitian positive definite and a $\lambda \in (0,1)$ and look at the convex combination $\lambda(A,\Gamma) + (1-\lambda)(B,\Phi)$ and the evaluation of your function. This leads to $$ (*) \quad \quad\|(\lambda\Gamma + (1-\lambda)\Phi) (\lambda A + (1-\lambda)B) (\lambda\Gamma + (1-\lambda)\Phi)^{-1}\|_F^2$$

With this in mind it is easy to construct a counterexample. Let $$A=\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}, \quad \Gamma=\begin{bmatrix} 1 & 0 \\ 0 & \varepsilon \end{bmatrix},\quad B=\begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}, \quad \Phi=\begin{bmatrix} \varepsilon & 0 \\ 0 & 1\end{bmatrix}.$$

Then $$\|\Gamma A\Gamma^{-1}\|_F^2 = \varepsilon^{-2} = \|\Phi B\Phi^{-1}\|_F^2$$ and for $\lambda =1/2$ plugged into (*) (note that one factor $1/2$ cancels) $$ \|1/2(\Gamma + \Phi) (A + B) (\Gamma + \Phi)^{-1}\|_F^2 = \frac{1}{4}\|A + B\|_F^2 = \frac{1}{2}.$$

Now choose any $\varepsilon > \sqrt{2}$ and you have a counterexample.