Consider a triangular array of rowise independent random variables $(X_{n,1}, \ldots,X_{n,n})_{n\geq 1}$ and assume $\mu_n=f_n(X_{n,1},\ldots,X_{n,n})$, for some measurable map $f_n:\mathbb{R}^n \to \mathbb{R}$. In particular, assume that, almost surely, $\mu_n\to c$, where $c$ is a finite constant.
For each integer $n \geq 1$ and $i=1, \ldots,n$, define $\bar{X}_{n,i}:=X_{n,i}-\mu_n$ and $\tilde{X}_{n,i}=X_{n,i}-c$; then let $$ \bar{X}_n:=(\bar{X}_{n,1},\ldots,\bar{X}_{n,n}),\\ \tilde{X}_n:=(\tilde{X}_{n,1},\ldots,\tilde{X}_{n,n}). $$
Question Is it true that the total variation distance $d_{TV}$ between $P(\bar{X}_n \in \cdot)$ and $P(\tilde{X}_n \in \cdot)$ converge to zero?
One may think of using the coupling inequality $$ d_{TV}(P(\bar{X}_n \in \cdot), P(\tilde{X}_n \in \cdot)) \leq 2P(\bar{X}_n\neq \tilde{X}_n ) \quad \textbf{(E.1)} $$ to establish $$ \lim_{n\to \infty}d_{TV}(P(\bar{X}_n \in \cdot), P(\tilde{X}_n \in \cdot)) =0. \hspace{4em}\textbf{(E.2)} $$
Though, that would not work: indeed, $P(\bar{X}_n\neq \tilde{X}_n )=P(\mu_n\neq c)$ and, as far as I understand, $\mu_n \to c$ almost surely does not entail $P(\mu_n\neq c)\to 0$, it only entails that $P(\lim_{n\to \infty}\mu_n= c)\to 0$. In fact, if $\mu_n$ has an absolutely continuous distribution, then we have $P(\mu_n\neq c)=1$, for every $n\geq 1$.
Is there another way to establish the result in (E.2), if true, eventually under additional conditions?
You essentially constructed your counter-example. Let $f_n = \mu_n$, and let $X_n$ be point masses. Two RVs who are point masses have TV 1 as long as their support isn't the same.
Even continuity alone would not help you; consider a sequence of RVs $X_n \rightarrow 0$ almost surely, and let $f_n(X_1, ..., X_n) = X_1$. Now, you are essentially asking for the TV between $X_n$ and $0$, which will be 1 if $X_n$ is absolutely continuous w.r.t. Lebesgue measure.