Equivalent markings on Riemann surfaces This is the definition in Introduction to Teichmuller Spaces by Imayoshi.
Let $(R,\Sigma_p)$ be a pair where $R$ is the riemann surface and $\Sigma_p$ is called marking where $\Sigma_p$ is a set of generator of $\pi_1(R,p)$. Two markings $\Sigma_p,\Sigma_q$ are equivalent if there is a curve inducing an isomorphism of $\pi_1(R,p)\cong\pi_1(R,q)$. 
Q: Shouldn't riemann surface path connected imply that all such markings are equivalent? Path connected says there is no difference in taking fundamental group. What am I missing here?
 A: Two markings are equivalent if they are related by conjugation (inner automorphism of $\pi_1(R)$). But there is still a big family of markings that are related by automorphisms of $\pi_1(R)$ that are not conjugations, which are non-equivalent. The group of automorphism quotiented by the inner automorphisms is called outer automorphism group of $\pi_1(R)$ and denoted $\operatorname{Out}(\pi_1(R))$.
For example, consider the markings $\Sigma_p = \{ [A_j], [B_j] \}$ and $\Sigma'_p = \{ {T_{B_1}}_*([A_j]), {T_{B_1}}_*([B_j]) \}$ where $T_{B_1} : R \to R$ is the Dehn twist about the curve $B_1$. Note that ${T_{B_1}}_*([A_1])=[A_1 \circ B_1]$, while ${T_{B_1}}_*([B_j])=[B_j]$ for $j=1,\cdots,g$ and ${T_{B_1}}_*([A_j])=[A_j]$ for $j=2,\cdots,g$, so these two markings are not related by a conjugation.
As a note, the group $\operatorname{Out}(\pi_1(R))$ is an extension of the mapping class group of the surface (see The Dehn–Nielsen–Baer theorem, a good reference on this is Primer on Mapping class groups by Farb and Margalit).
