If $(Y_n)$ is iid, then $Z_n:=\sum_{i=1}^nY_i$ is Markov

Question

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_n)_{n\in\mathbb N_0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$, $E$ be a $\mathbb R$-Banach space, $(Y_n)_{n\in\mathbb N}$ be an $E$-valued $(\mathcal F)_{n\in\mathbb N}$-adapted independent identically distributed process on $(\Omega,\mathcal A,\operatorname P)$ and$^1$ $$Z_n:=\sum_{i=1}^nY_i\;\;\;\text{for }n\in\mathbb N_0.$$

Can we conclude that $(Z_n)_{n\in\mathbb N_0}$ is a time-homogeneous $(\mathcal F_n)_{n\in\mathbb N_0}$-Markov chain?

Let $B\in\mathcal B(E)$. Then $$f(x,y):=1_B(x+y)\;\;\;\text{for }x,y\in E$$ is bounded and Borel measurable. Let $n\in\mathbb N_0$. Since $\mathcal F_n$ and $Y_{n+1}\sim Y_1$ are independent and $Z_n$ is $\mathcal F_n$-measurable, we should have \begin{equation}\begin{split}\operatorname P\left[Z_{n+1}\in B\mid\mathcal F_n\right]&=\operatorname E\left[f(Z_n,Y_{n+1})\mid\mathcal F_n\right]=\left.\operatorname E\left[f(x,Y_{n+1})\right]\right|_{x\:=\:Z_n}\\&=\left.\operatorname P\left[Y_{n+1}\in B-x\right]\right|_{x\:=\:Z_n}=\left.\operatorname P\left[Y_1\in B-x\right]\right|_{x\:=\:Z_n}.\end{split}\tag1\end{equation}

There are two things I worry about:

1. I think we need to assume that $Y_1$ is independent of $\mathcal F_0$ or, maybe better, to simply assume that the sequence contains a $Y_0$.

2. Moreover, I think the assumption of $\mathcal F$-adaptedness and independence is not enough to ensure that $Y_{n+1}$ is independent of $\mathcal F_n$, since we seem to know only that $Y_{n+1}$ is independent of $\sigma(Y_1,\ldots,Y_n)\subseteq\mathcal F_n$. Can we fix this?

BTW, the transition kernel of $(Z_n)_{n\in\mathbb N_0}$ should be $$\kappa(x,B):=\mu(B-x)\;\;\;\text{for }(x,B)\in E\times\mathcal B(E)\tag2,$$ where $\mu:=\mathcal L(Y_1)$.

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$^1$ $Z_0:=0$.