question about fiber functors and fundamental groups I apologize for the vague question title, but I can't think of a better subject line.
Let $f : X\rightarrow S$ be a morphism of connected schemes. Let $x,y\in X$, $s\in S$ be geometric points with $f(x) = f(y) = s$. 
Let $FEt_S$ be the category of finite etale schemes over $S$, and similarly with $X$. We have fiber functors $F_s : FEt_S\rightarrow Sets$, and $F_x,F_y : FEt_X\rightarrow Sets$, and $\pi_1(X,x),\pi_1(X,y),\pi_1(S,s)$ are just the automorphism groups of the associated fiber functors.
There are canonical isomorphisms
$$\beta_x : F_x\circ f^*\stackrel{\sim}{\longrightarrow}F_s\qquad\text{and}\qquad \beta_y : F_y\circ f^* \stackrel{\sim}{\longrightarrow}F_s$$
Any isomorphism $\alpha : F_x\stackrel{\sim}{\longrightarrow} F_y$ induces an automorphism $$\overline{\alpha} := \beta_y\circ\alpha\circ\beta_x^{-1}\in Aut(F_s) = \pi_1(S,s)$$
Suppose the induced map $f_* : \pi_1(X,x)\rightarrow \pi_1(S,s)$ is trivial (sends everything to the identity), and similarly $f_* : \pi_1(X,y)\rightarrow\pi_1(S,s)$, then I believe I can prove that $\overline{\alpha}$ does not depend the choice of $\alpha$. However, I'm having a lot of trouble arguing that $\overline{\alpha}$ should always be $id\in Aut(F_s)$.
Is this even true? (I mean, surely it is right. "What else could it be?") How would one prove something like this?
EDIT 1: The specific case I'm interested in is where $X\rightarrow S$ actually factors through the geometric point $s$.
EDIT 2: Okay it seems that this additional assumption makes the statement pretty easy to prove, though I would be interested in a counterexample to the original question.
 A: I don't think that $\bar\alpha$ is always the identity. You should think of $\alpha$ as a ``path'' from $x$ to $y$, so in principle it is possible that the loop obtained as the image of $\alpha$ is not nullhomotopic, even if the image of every loop is. In fact, if the map $f$ is the universal covering space you can get every possible element in $\pi_1(S,s)$ by choosing appropriately $x$ and $y$. To support this let me try to get you an example where $\bar \alpha$ is not in the image of $f_*$ (although $f_*$ is not trivial). 
Fix the ground field $\mathbb{C}$. Take $f:\mathbb{A}^1\smallsetminus \{0\}\to \mathbb{A}^1\smallsetminus \{0\}$ be the square map, and take $x=1$ and $y=-1$. Let the isomorphism $\alpha: F_x\xrightarrow{\sim} F_y$ to be obtained by "sliding" the points of the fiber along the half unit circle in the positive semiplane. Then $\bar\alpha$ should be the generator of $\pi_1(\mathbb{A}^1\smallsetminus\{0\})=\hat{\mathbb{Z}}$, while the image of $f_*$ is the index 2 subgroup.
