Consider the following bayesian network depicted below
If $X_{1:L}$ are i.i.d. random variables where, for any $i \in [1:L]$, $X_i \in \mathcal{X}$ is uniformly distributed, we can factorize $q^{(1)}_{Z_1\dots Z_L}$ as
\begin{equation}
q^{(1)}_{Z_1\dots Z_L} = \frac{1}{|\mathcal{X}|^L} \sum_{x_{1:L}} \prod_{i=1}^L p_{Z|AB}(z_i|x_i,x_{i-1}).
\end{equation}
Now, consider a new bayesian network, where, for all $i \in [1:L]$, the dependency between $X_{i-1}$ and $Z_{i}$ is removed, and $X_{i-1}$ is replaced by a new r.v. $\bar{X}_{i-1}$ equally distributed and connected to $Z_{i}$. In this case, we have \begin{align} q^{(2)}_{Z_1\dots Z_L} & = \frac{1}{|\mathcal{X}|^{2L}} \sum_{x_{1:L}}\prod_{i=1}^L \sum_{\bar{x}_{i-1}} p_{Z|AB}(z_i|x_i,\bar{x}_{i-1}) \end{align}
The total variation distance between $q^{(1)}_{Z_1\dots Z_L}$ and $q^{(2)}_{Z_1\dots Z_L}$ is \begin{align} & d_{TV}(q^{(2)}_{Z_1\dots Z_L},q^{(1)}_{Z_1\dots Z_L}) \\ & \quad = \sum_{z_{1:L}} \bigg| q^{(2)}_{Z_1\dots Z_L}(z_{1:L}) - q^{(1)}_{Z_1\dots Z_L}(z_{1:L}) \bigg| \\ & \quad = \sum_{z_{1:L}} \bigg| \frac{1}{|\mathcal{X}|^{2L}} \sum_{x_{1:L}} \prod_{i=1}^L \sum_{\bar{x}_{i-1}} p_{Z|AB}(z_i|x_i,\bar{x}_{i-1}) - \frac{1}{|\mathcal{X}|^L} \sum_{x_{1:L}} \prod_{i=1}^L p_{Z|AB}(z_i|x_i,x_{i-1}) \bigg| \end{align}
If $p_{Z|AB}(z|a,b) = p_{Z|AB}(z|a,b^{\prime})$ for any $a,b,b^{\prime} \in \mathcal{X}$ (that is, $Z$ is determined only by $A$), we have \begin{align} & d_{TV}(q^{(2)}_{Z_1\dots Z_L},q^{(1)}_{Z_1\dots Z_L}) \\ & \quad = \sum_{z_{1:L}} \bigg| \frac{1}{|\mathcal{X}|^{2L}} \sum_{x_{1:L}} \prod_{i=1}^L |\mathcal{X}| p_{Z|AB}(z_i|x_i,x^{\prime}) - \frac{1}{|\mathcal{X}|^L} \sum_{x_{1:L}} \prod_{i=1}^L p_{Z|AB}(z_i|x_i,x^{\prime}) \bigg| \\ & \quad = \sum_{z_{1:L}} \bigg| \frac{1}{|\mathcal{X}|^{L}} \sum_{x_{1:L}} \prod_{i=1}^L p_{Z|AB}(z_i|x_i,x^{\prime}) - \frac{1}{|\mathcal{X}|^L} \sum_{x_{1:L}} \prod_{i=1}^L p_{Z|AB}(z_i|x_i,x^{\prime}) \bigg| \\ & \quad = 0. \end{align}
QUESTION: Is it also true that $d_{TV}(q^{(2)}_{Z_1\dots Z_L},q^{(1)}_{Z_1\dots Z_L}) = 0$ implies $p_{Z|AB}(z|a,b) = p_{Z|AB}(z|a,b^{\prime})$ for any $a,b,b^{\prime} \in \mathcal{X}$.