Let $(X_n)_{n\geq1}$ be an i.i.d. sequence with $\mathbb{E}\{|X_1|\}<\infty$. Let $S_n=X_1+\cdots+X_n$ and $\mathcal{F}_{-n}=\sigma(S_n,S_{n+1},\ldots)$. Then, one can state that $$M_{-n}=\mathbb{E}\{X_1\mid \mathcal{F}_{-n}\}$$ is a backwards martingale and, by Law of Iterated Expectation, $\mathbb{E}\{M_{-n}\}=\mathbb{E}\{X_1\}$.
At this point, how can one prove that
by symmetry for $1\leq j\leq n$:$$\mathbb{E}\{X_1\mid \mathcal{F}_{-n}\}=\mathbb{E}\{X_j \mid \mathcal{F}_{-n}\}\hspace{0.2cm} \text{ a.s.}$$
Hint: Define a permutation $f$ on $\mathbb N$ by $f(1)=j, f(j)=1$ and $f(k)=k$ for $k \notin \{1,j\}$. Let $Y_n=X_{f(n)}$. Then $(X_n)$ and $(Y_n)$ are identically distributed. This implies that $\mathbb E (X_1|\mathcal F_{-n})=E (X_j|\mathcal F_{-n})$ because for any $E \in \mathcal F_{-n}$ we have $\mathbb EX_1 1_E=\mathbb EY_1 1_E=\mathbb EX_j 1_E$. [Note that $\mathcal F_{-n}$ does not change under the permutation $f$].