If $(Y_{i})_{\{i = 1,\ldots,T\} }$ is an IID sequence, what equivalencies do I get with the conditional expectation on $\sigma \left(\sum\limits_{i=t+1}^n Y_i\right)$ where $t \in \{1,...,T\}$. It is intuitive - by virtue of $Y$ being IID that for all $A:=\left\{ \sum\limits_{i=t+1}^n Y_i \in B\right\}$, where $B \in \mathcal{B}(\mathbb R)$, we get:
$E[Y_{t+1} 1_A]=E[Y_{t+2}1_A]=\cdots=E[Y_T 1_A]$ which then of course implies that
$$ E\left[Y_{t+1}\mid \sigma\left(\sum\limits_{i=t+1}^n Y_i\right)\right] = E\left[Y_{t+2}\mid \sigma\left(\sum\limits_{i=t+1}^n Y_i\right)\right] = \cdots = E\left[Y_T \mid \sigma\left(\sum\limits_{i=t+1}^n Y_i\right)\right] $$
My question regards any $Y_{j}$ where $j \in \{1,\ldots,t\}$. Surely by virtue of being IID with the other $(Y_i)_{i=t+1,\ldots,T}$ would imply that $$ E\left[Y_j \mid \sigma\left(\sum_{k=t+1}^n Y_k\right)\right] = E\left[Y_i \mid \sigma\left(\sum_{k=t+1}^n Y_k\right)\right] \tag{$*$} $$ The only argument that could go against my assertion would be the fact that $Y_j$ is independent of $\sigma \left( \sum\limits_{k=t+1}^n Y_k \right)$.
My basic misunderstanding: Since $Y_{i}$ and $Y_{j}$ have the same distribution then $\sigma (Y_i) =\sigma (Y_j)$ (i.e. $Y_i$ and $Y_j$ describe the same measurable sets) and hence from that point of view, shouldn't $(*)$ hold?
Indeed, your computation showed that for all $j\in\{t+1,\dots,n\}$, $$ \mathbb E\left[Y_j\mid \sigma\left(\sum_{k=t+1}^nY_k\right)\right]=\frac 1{n-t}\sum_{k=t+1}^nY_k. $$ If $j\in\{1,\dots,t\}$, then $Y_j$ is independent of $Y_{t+1},\dots,Y_n$ hence $Y_j$ is independent of $\sigma\left(\sum_{k=t+1}^nY_k\right)$ which implies $$ \mathbb E\left[Y_j\mid \sigma\left(\sum_{k=t+1}^nY_k\right)\right]=\mathbb E\left[Y_1\right]. $$