That would be mostly kinda correct if X was assumed symmetric, for then you could turn XB into $\left({B}^{t}{X}^{t}\right)}^{t}={\left({B}^{t}{X}^{t}\right)}^{t$. You'd still have an extra transpose to get rid of, though, and for that, you'd need to also assume that ${B}^{t}X$ was symmetric.

So...the two systems are very far from being equivalent. You might want to try this with, say

$A=((2\text{}\text{}0),(0\text{}\text{}1))$

$B=((1\text{}\text{}2),(0\text{}\text{}1))$

and

$C=((3\text{}\text{}1),(0\text{}\text{}3))$

Write out your version of things, solve for X, and plug it into the original: it won't be a solution.