$\newcommand{\eD}{\overset{\text{D}}\to}$
Let $X_n$ and $X$ be any random vectors with distributions $\mu_n$ and $\mu$, respectively. Let $Y_n$ be an independent copy of $X_n$, for each $n$, and let $Y$ be an independent copy of $X$. The questions can then be restated as follows:
Q1: Does $X_n\eD X$ imply $X_n-Y_n\eD X-Y$?
Q2: Vice versa, does $X_n-Y_n\eD X-Y$ imply $X_n\eD X$?
Here $\eD$ denotes the convergence in distribution.
The answer to Q1 is yes. Indeed, let $g_Z$ denote the characteristic function (c.f.) of a random vector $Z$. Then $X_n\eD X$ means that $g_{X_n}\to g_X$ pointwise, whence $g_{X_n-Y_n}=|g_{X_n}|^2\to|g_X|^2=g_{X-Y}$, so that $X_n-Y_n\eD X-Y$.
The answer to Q2 is no. Indeed, let $X_n=n=Y_n$ and $X=Y=0$. Then $X_n-Y_n\eD X-Y$, but $X_n\eD X$ is false.