Given that $A^T=A,B^T=-B$

Now solving all option one by one we get,

Solving $\left(A\right)$ option 

$C=A^4-B^4$

$C^T=\left(A^4-B^4\right)={\left(A^4\right)}^T-{\left(B^4\right)}^T=A^4-B^4=C$

$\Rightarrow C=C^T$ is a symmetric matrix

Solving $\left(B\right)$ option 

$C=AB-BA$

$C^T={\left(AB-BA\right)}^T={\left(AB\right)}^T-{\left(BA\right)}^T$

$=B^T{A}^T-A^T{B}^T=-BA+AB=C$

$\Rightarrow C=C^T$ is a symmetric matrix

Solving $\left(C\right)$ option 

$C=B^5-A^5$

$C^T={\left(B^5-A^5\right)}^T={\left(B^5\right)}^T-{\left(A^5\right)}^T=-B^5-A^5$

$\Rightarrow C\ne C^T \text{or} C\ne -C^T$

Solving $\left(D\right)$ option 

$C=AB+BA$

$C^T={\left(AB+BA\right)}^T={\left(AB\right)}^T+{\left(BA\right)}^T$

$=-BA-AB=-C$

$C=-C^T$ is skew symmetric matrix

$\therefore$ Option (C) is not true.