Central limit theorem replacing radical n with n Does $n(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2/ n^2),$ remembering that $n^{1/2}(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2)?$

No, write
$$n(y_n-\mu)=n^{1/2}n^{1/2}(y_n-\mu)$$
From $n^{1/2}(y_n-\mu) \overset{\mathbb d}{\to} \mathcal N(0,\sigma^2)=O_p(1)$ (cf. e.g. root-n consistent estimator, but root-n doesn't converge?) we find $$n(y_n-\mu)=O_p(n^{1/2})$$ we obtain something which diverges whereas, as Nicolas has pointed out, $N(0,\sigma^2/ n^2)$ tends to a degenerate r.v. at zero, so basically the constant zero.
