From page 114 of Rotman's Algebraic Topology:
If $X$ is a space, then the $n$-th Barycentric Subdivision is defined as the homomorphism $Sd_n : S_n(X) \rightarrow S_n(X)$ defined on generators $\Delta^n \rightarrow X$ by $Sd_n(\sigma) = \sigma_\text{#}Sd_n(\delta^n)$, where $\sigma_{\text{#}} : S_n(\Delta^n) \rightarrow S_n(X)$ and $\delta^n: \Delta^n \rightarrow \Delta^n$ is the identity mapping.
I don't understand how this mapping works, it almost seems like it's just the identity map. If $\sigma_1, \sigma_2$ are two arbitrary generators of $X$, then $Sd_n(\sigma_1) = (\sigma_1)_{\text{#}}Sd_n(\delta^n)$ and $Sd_n(\sigma_2) = (\sigma_2)_{\text{#}}Sd_n(\delta^n)$. But if $\delta^n : \Delta^n \rightarrow \Delta^n$ is the identity map and both $\sigma_{1,2} : \Delta^n \rightarrow X$, then wouldn't they both map to the same element of $S_n(X)$?
When you substitute $\sigma=\sigma_1$ into the formula $$Sd_n(\sigma)=\sigma_\#(Sd_n(\delta^n)) $$ you get $$Sd_n(\sigma_1)=(\sigma_1)_\#(Sd_n(\delta^n)) $$ not $$Sd_n(\sigma_1)=\sigma_\#(Sd_n(\delta^n)) $$