Is there a matrix $C$ that can make $AB$ return the same result as $BA$ where $C$ is based on $B$?

To signify that $C$ only depends on $B,$ let $C=f(B).$ This gives us $BA=ABf(B)$ for all $A.$ Plugging in $A=I,$ we get $f(B)=I.$ Thus, $BA=AB,$ which means $B$ commutes with every matrix. This is only possible if $B$ is a multiple of the identity, and in this case $C=I$ indeed works.