Definition: A marking on a sphere $S$ with $g$ handles ($g<\infty$) is an ordered collection $\Sigma_p=\{[\alpha_j],[\beta_j]\}_{j=1}^g$ of elements $[\alpha_j],[\beta_j]$ all of them in a same fundamental group $\pi_1(p,S),p\in S$ such that $\Sigma_p$ generates $\pi_1(p,S)$ and $\prod\limits_{j=1}^g[[\alpha_j],[\beta_j]]=1$, where $[[\alpha_j],[\beta_j]]$ denotes the comutator of $[\alpha_j]$ and $[\beta_j]$ (namely, $[\alpha_j][\beta_j][\alpha_j]^{-1}[\beta_j]^{-1}$).
Are the answers to the following questions "Yes"? (I wish they were, cause then I could use them in the proof of a theorem...):
Question 1: Given two markings $\Sigma_p=\{[\alpha_j],[\beta_j]\}_{j=1}^g$ and $\Sigma'_p=\{[\alpha_j],[\beta_j]\}_{j=1}^g$ on a sphere $S$ with $g$ handles does the equations $$F([\alpha_j]):=[\alpha'_j],\,F([\beta_j]):=[\beta'_j], \, j=1,\dots,g$$ (which we could denote as $F(\Sigma_p)=\Sigma'_p$) define an automorphism $F:\pi_1(p,S)\to \pi_1(p,S)$?
The idea of question 1 is to create an isomorphism using the "send-basis-to-basis" process, but the problem is that $\Sigma_p$ and $\Sigma_p'$ are not really basis, since an element $[\alpha]\in \pi_1(p,S)$ might have not a unique representation in terms of elements of $\Sigma_p$ (or $\Sigma'_p$).
Question 2: If the answer to question 1 is "Yes", is it true that $F\in \mathrm{Inn}(\pi_1(p,S))$? I mean, there exist a $[\gamma]\in \pi_1(p,S)$ such that $$F([\alpha])=[\gamma][\alpha][\gamma]^{-1},\,\forall [\alpha]\in \pi_1(p,S)?$$
Any proofs or counterexamples?
EDIT: I think I've made some progress. Such an $F$ as defined in question 1 is an isomorphism.
If the genus $g$ is zero (the sphere case), the problem of question 2 is trivial (since $\pi_1(p,S)=\{e\}$). In the case $g=1$ (torus) question 2 is false! In fact, let $\Sigma_p=\{[\alpha_1],[\beta_1]\,:\, [\alpha_1][\beta_1][\alpha_1]^{-1}[\beta_1]^{-1}=1\}$ be any marking with basepoint $p$ on $S$. So $\Sigma'_p=\{[\alpha'_1],[\beta'_1]\}$ with $[\alpha'_1]:=[\alpha_1]$ and $[\beta'_1]=[\alpha_1][\beta_1]$ is a marking on $S$ with basepoint $p$. In fact, $$\begin{array}{rcl}\Pi_{i=1}^1[[\alpha'_i],[\beta'_i]]&=&[\alpha'_1][\beta'_1][\alpha'_1]^{-1}[\beta'_1]^{-1}\\ &=&[\alpha_1][\alpha_1][\beta_1][\alpha_1]^{-1}[\beta_1]^{-1}[\alpha_1]^{-1}\\ &=&1, \end{array}$$ and $[\alpha_1]=[\alpha_1']$, $[\beta_1]=[\beta'_1][\alpha'_1]^{-1}$ shows that $\langle \Sigma'_p\rangle=\pi_1(p,S)$.
Although, there is no $[\gamma]\in \pi_1(p,S)$ such that $F([\alpha])=[\gamma][\alpha][\gamma]^{-1},\forall [\alpha]\in \pi_1(p,S)$ because, since $\pi_1(p,S)$ is abelian, $[\gamma][\alpha][\gamma]^{-1}=[\alpha],\forall [\alpha],[\gamma]\in \pi_1(p,S)$, but $F\neq Id:\pi_1(p,S)\to \pi_1(p,S)$, because $F([\beta_1])\neq [\beta_1]$. So...
New Question: Is there any chances of the answer to question 2 to be "Yes", for $g\geq 2$? (Since $\pi_1(p,S)$ is not abelian for in these cases...)
Question 2 has negative answer. There is an important group describing this failure, called, the (extended) mapping class group (MCG) of $S$, it is isomorphic to $Out(\pi_1(S))=Aut(\pi_1(S))/Inn(\pi_1(S))$, where $Inn$ denotes the subgroup of inner automorphisms. See: https://en.wikipedia.org/wiki/Mapping_class_group_of_a_surface#The_Dehn.E2.80.93Nielsen.E2.80.93Baer_theorem and references therein. Examples of nontrivial elements of MCG abound, for instance, all Dehn twists are such.