Here is a way to change the question to make it easier to answer.

Let $p:X \to B$ be a topological fiber bundle with fiber $F$, such that $F$, $X$, and $B$ are all connected and tame (say, CW complexes). Then there is a homotopy exact sequence
$$\pi_2(B) \to \pi_1(F) \to \pi_1(X) \to \pi_1(B) \to 1.$$
Let $N$ be the quotient of $\pi_1(F)$ by the image of $\pi_2(X)$. Then $N$ is also a subgroup of $\pi_1(X)$, and
$$\pi_1(X)/N \cong \pi_1(B).$$
So $\pi_1(B)$ acts on $N$ because a quotient group always acts by conjugation on the kernel.

If $X$ has a global section as a bundle over $B$, this implies that the term $\pi_2(B) \to \pi_1(F)$ is trivial (because a sphere in $\pi_2(B)$ then lifts to a disk mapped to $X$ whose boundary is a constant loop in $F$). If this term is trivial, whether or not $X$ has a global section, then $\pi_1(F) = N$ and so $\pi_1(B)$ acts on $\pi_1(F)$.

Now, suppose that $p:X \to B$ is a map of algebraic varieties and $F$ is some fiber over a closed point. Then the étale fundamental group $\tilde{\pi}_1$ is a functor on the relevant category. So you at least get a sequence
$$\tilde{\pi}_1(F) \to \tilde{\pi}_1(X) \to \tilde{\pi}_1(B) \to 1.$$
Then maybe the composition of the first two maps is trivial because $F$ is sent to a closed point. For a general $p$ (taking simple examples from topology), this sequence is not an exact sequence. But, if there are suitable conditions on $p$ for it to be analogous to a fiber bundle, then maybe it should be exact. Note that the étale fundamental group $\tilde{\pi}_1(V)$ is the profinite completion of the analytic fundamental group $\pi_1(V)$ when $V$ is a variety over $\mathbb{C}$. On the negative side, profinite completion is not an exact functor on groups. (But what about for fundamental groups of complex algebraic varieties?)

If it is exact, then I'm not sure whether the kernel in $\tilde{\pi}_1(F)$ can be naturally related to anything like an étale $\tilde{\pi}_2(B)$, but maybe so.