Intuition behind commutativity of convolution in LTI systems
In a discrete-time system such as the one that you have, the number $y[n_0]$ (here $n_0$ is a fixed integer) is a sum of the form $$\sum_{k=-\infty}^\infty h[k]x[n_0-k]$$ which can be re-arranged via a change of variables (replace $k$ by $n_0-\ell$) to $$\sum_{\ell=-\infty}^\infty h[n_0-\ell]x[\ell].$$ So, the commutativity of the convolution is trivial.  The issue is the interpretation that you put on it.  As Laurent Duval/s answer points out, the systems A and B are not equivalent in any sense of the term.  If the signal $x$ were replaced by a different signal $\hat{x}$, then system A would have output $\hat{y} = h \star \hat{x}$, but you wouldn't get the same output $\hat{y}$ if system B were excited by $h$; the impulse response of system B continues to be $x$, and system B thus has output $x \star \hat{x} =  \hat{x}\star x \neq h \star \hat{x}$.