Show such finite sequence of random variables is exchangable

Given random variables $$X_1$$,$$X_2$$,...,$$X_n$$ i.i.d, $$Z$$ is random variable which is independent of $$X_1$$,$$X_2$$,...,$$X_n$$, Let $$Y_i = X_i + Z, i = 1,2,...,n$$, are $$Y_1, Y_2, ... , Y_n$$ exchangable(I guess so)?If so, how to show it?If not, what about $$Z$$ is also iid with $$X_1$$,$$X_2$$,...,$$X_n$$?

One way is to use characteristic functions. We have $$Ee^{i\sum_j t_jY_j}=\prod_jEe^{it_jX_j} Ee^{i \sum_j t_jZ}$$ which clearly doesn't change under any permutation of $$t_j$$'s. It is not necessary to assume that $$Z$$ has the same distribution as $$X_i$$'s.
Let $$X,Z\colon (\Omega,\Sigma,\mathbb P)\to \mathbb R$$ be two random variables and $$X_1,\dots,X_n\stackrel{\rm i.i.d.}{\sim} X$$ be i.i.d. random variables that are independent of $$Z$$. For $$A\in\mathcal B(\mathbb R)$$, let $$A-z := \{a-z|a\in A\}$$ denote the shift of the set $$A$$ by $$z\in\mathbb R$$.
It is easier to first think of $$Z$$ as being a discrete random variable. Then for $$A_1,\dots,A_n\in\mathcal B(\mathbb R)$$ we obtain (by the law of total probability and the independence of all random variables: \begin{align*} \mathbb P\big( Y_1\in A_1,\dots, Y_n\in A_n\big) &= \sum_z \mathbb P(Z=z) \mathbb P\big( X_1\in A_1-z,\dots, X_n\in A_n-z\big) \\ &= \sum_z \mathbb P(Z=z) \mathbb P\big( X\in (A_1-z)\big)\cdots \mathbb P\big( X\in (A_n-z)\big). \end{align*} Since $$X_1,\dots,X_n$$ are i.i.d., you obtain the same result for $$\mathbb P\big( Y_{\sigma(1)}\in A_1,\dots, Y_{\sigma(n)}\in A_n\big)$$, where $$\sigma$$ is some permutation of $$\{1,\dots,n\}$$.
If $$Z$$ is not discrete, we have to use integrals instead of sums: \begin{align*} \mathbb P\big( Y_1\in A_1,\dots, Y_n\in A_n\big) &= \int_\mathbb R \mathbb P\big( X_1\in A_1-z,\dots, X_n\in A_n-z\big) \, \mathrm d\mathbb P_Z(z) \\ &= \int_\mathbb R \mathbb P\big( X\in (A_1-z)\big)\cdots \mathbb P\big( X\in (A_n-z)\big) \, \mathrm d\mathbb P_Z(z), \end{align*} where $$\mathbb P_Z$$ is the distribution of $$Z$$.
1. It is not necessary for $$X$$ and $$Z$$ to map into $$\mathbb R$$. It can be any space you like. In order to make your question well-posed, you have to be able to perform the addition $$X+Z$$, though, so probably they map into the same vector space $$V$$.
2. We used the fact that the product $$\sigma$$-algebra is generated by the sets of the form $$A_1\times\cdots\times A_n$$.