TITLE: Copies of topological fundamental groups inside etale fundamental groups given by different embeddings of your field into $\mathbb{C}$
QUESTION [10 upvotes]: Let $X$ be a smooth curve over a number field $K$ (not necessarily proper). Fix an algebraic closure $\overline{K}$ of $K$.
Let $i,i' : \overline{K}\hookrightarrow\mathbb{C}$ be two abstract embeddings (ie, as $\mathbb{Q}$-algebras). Let $X_\mathbb{C},X_{\mathbb{C}}'$ be the base changes of $X_{\overline{K}}$ to $\mathbb{C}$ via $i$ and $i'$. Then, $X_\mathbb{C}(\mathbb{C})$ has the structure of a Riemann surface. Let $x\in X_\mathbb{C}(\mathbb{C})$ come from a $\overline{K}$-rational point. We may consider its topological fundamental group $\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)$. Since every finite cover of the Riemann surface $X_\mathbb{C}(\mathbb{C})$ is algebraic, for every loop in $\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)$, its monodromy action on the fibers at $x$ of its finite covers determines an automorphism of the fiber functor at $x$, and hence we obtain a homomorphism
$$\pi_1^{top}(X_\mathbb{C}(\mathbb{C}),x)\rightarrow \pi_1^{et}(X_\mathbb{C},x)$$
which is known to be the embedding of the first group into its profinite completion. The map $X_\mathbb{C}\rightarrow X_{\overline{K}}$ given by base change induces an isomorphism on etale fundamental groups, and composing these maps we get
$$\pi_1^{top}(X_\mathbb{C}(\mathbb{C}))\longrightarrow \pi_1^{et}(X_\mathbb{C})\stackrel{\sim}{\longrightarrow}\pi_1^{et}(X_\overline{K})$$
where I've omitted the base points because I only care about these maps up to conjugacy (say, inside $\pi_1^{et}(X_{\overline{K}})$). Similarly, with $X_\mathbb{C}'$, we get a map
$$\pi_1^{top}(X'_\mathbb{C}(\mathbb{C}))\longrightarrow \pi_1^{et}(X'_\mathbb{C})\stackrel{\sim}{\longrightarrow}\pi_1^{et}(X_\overline{K})$$
Both of these maps give embeddings of the topological fundamental groups inside $\pi_1^{et}(X_{\overline{K}})$, canonical up to conjugation.
My question is:
When are the images the same (up to conjugation)?
Are there examples when the images are not the same?
I'm particularly interested in the case when $X_\mathbb{C}(\mathbb{C})$ is hyperbolic.
References would also be appreciated.

REPLY [2 votes]: For affine hyperbolic curves, when $i$ and $i'$ agree on $K$, this happens only when $i$ and $i'$ are equal or complex conjugate of each other.
Let $f: \pi_1^{top}(X) \to \pi_1^{et}(X_{\mathbb C})$ be the natural dense inclusion, and $e_i,e_{i'} \pi_1^{et}(X_{\mathbb C}) \to \pi_1^{et}(X_{\overline{K}})$ be the natural isomorphisms defined by $i$ and $i'$. Let $\sigma$ be the element of the Galois group of $\overline{K}$ over $K$ that sends $i$ to $i'$, then $e_{i'}$ is $e_i$ composed with the action of $\sigma$ by outer automorphism of $\pi_1^{et}(X_{\overline{K}})$.
If the image of $e_i \circ f$ is conjugate to $e_{i'} \circ f$, then there must be some automorphism $\alpha$ of $\pi_1^{top}(X)$ such that $e_i \circ f \circ \alpha$ is conjugate  to $e_{i'} \circ f = \sigma \circ e_i \circ f$ as a homomorphism.  Now because $f$ is dense, we can extend $\alpha$ to an outer automorphism of $\pi_1^{et}(X_{\mathbb C})$ and thus to an outer automorphism of $\pi_1^{et}(X_{\overline{K}})$, and then the condition that these two maps are conjugate becomes the condition that these two outer automorphisms are equal in the outer automorphism group.
It is a result of Matsumoto and Tamagawa that this can only happen for the identity and comple conjugation (Mapping-Class-Group Action versus Galois Action on Profinite Fundamental Groups, Remark 2.1).
Furthermore, it looks to me that following their argument, the only case that the affineness assumption is used can be replaced with Theorem C(i) of Hoshi and Mochizuki