Action on étale fundamental group is conjugation Let $X$ be a quasi-compact, quasi-separated connected scheme and let $\bar{x}$ be a geometric point. Denote by $\pi_1(X,\bar{x})$ the étale fundamental group, defined as the automorphism of the fiber functor $$\bar{x}: \mathrm{FÉt}_X \to \mathrm{Set},$$ where $\mathrm{FÉt}_X$ is the category of schemes finite étale over $X$.
Suppose now that I have an element $\sigma \in \pi_1(X,\bar{x})$ and suppose that $\sigma$ lies in the center. Then for any $Y/X$ finite étale, $\sigma$ induces an automorphism of $Y$ over $X$, which we call $\sigma_Y$.
Since $\sigma_Y$ is an automorphism of $Y$, we get a map $$(\sigma_Y)_\ast: \pi_1(Y,\bar{y}) \to \pi_1(Y,\sigma_y(\bar{y})).$$ Composing with the choice of a path $\gamma:\bar{y} \to \sigma_y(\bar{y})$, we get an automorphism
$$(\sigma'_Y)_\ast = \gamma_\ast \circ (\sigma_Y)_\ast : \pi_1(Y,\bar{y}) \to \pi_1(Y,\bar{y}).$$
I have seen it claimed in Curves and their fundamental groups pg. 133, proof of Lemma, by Faltings that the automorphism $(\sigma'_Y)_\ast$ is inner. I don't see why this is true, could someone help explain the argument?
 A: Here is a potential answer to the question, but I am not sure whether it is correct, so to whoever reads this answer, please tell me if you find an error in my reasoning.
By assumption, $\sigma$ lies in the center. In particular, if $H = \pi_1(Y,\bar{y})$ is the (open) subgroup of $ G = \pi_1(X,\bar{x})$ corresponding to $H$ and $K \subset H$ is an open subgroup, the diagram
$$\require{AMScd}
\begin{CD}
G/K @>{R_{\sigma}}>> G/K\\
@VVV @VVV \\
G/H @>{R_{\sigma}}>> G/H
\end{CD}$$
of finite $G$-sets commutes, where $R_\sigma$ is right-multiplication with $\sigma$, and where the vertical maps are just the projection. Since the category of $G$-sets is equivalent to the category of schemes finite étale over $X$, we should be able to deduce that for $Z$ finite étale over $Y$, we have a commutative diagram
$$\require{AMScd}
\begin{CD}
Z @>{\sigma_Z}>> Z\\
@VVV @VVV \\
Y @>{\sigma_y}>> Y
\end{CD}$$
where $\sigma_y$ and $\sigma_z$ are the maps corresponding to $R_\sigma$.  Note that this diagram is a pull-back as well.
For any $Z/Y$ étale, we have a commutative diagram of cospans
$$\require{AMScd}
\begin{CD}
\bar{y} @>{j}>> Y  @<<{p_Z}< Z \\
@VV\mathrm{id}V @V\sigma_YVV @VV \sigma_Z V  \\
\bar{y} @>{\sigma_Y \circ j}>> Y @<<{p_Z}< Z 
\end{CD}$$
and by functoriality of pull-back, we get an induced map $Z_{\bar{y}} \to Z_{\sigma(\bar{y})}$, which defines a natural transformation
$F_{\bar{y}} \to F_{\sigma({\bar{y}})}$.
$\DeclareMathOperator\Aut{Aut}$The induced map $\sigma_y: \pi_1(Y,\bar{y}) = \Aut(F_\bar{y}) \to \Aut(F_{\sigma(\bar{y})}) = \pi_1(Y,\sigma(\bar{y}))$ is now defined, up to an inner homomorphism of $\pi_1(Y,\sigma(\bar{y})$, as follows. We take a natural automorphism $(\tau_Z)_{Z \in \mathrm{FÉt}_Y}$ and map it to the natural automorphism $(\tau'_Z)_{Z \in \mathrm{FÉt}_Y}$ of $F_{\sigma(\bar{y})}$, where $\tau'_Z = \tau_{\sigma(Z)}$. Here $\sigma(Z)$ is the finite étale scheme over $Y$ defined as $Z \to Y \xrightarrow{\sigma_y} Y$, where the first map is the "original" projection of $Z$ to $Y$. One notes that $\tau'_{\sigma(Z)} = \sigma_Z \tau_Z \sigma_Z^{-1}$, where $\sigma_Z$ is the map $Z_{\bar{y}} \to Z_{\sigma(\bar{y})}$ induced by $\sigma_Z: Z \to Z$ on fibers.
Now, on the other hand, we wish to compute the map
$$\pi_1(Y,\sigma(\bar{y})) \to \pi_1(Y,\bar{y})$$ induced by an étale path; we will see that there is a particularly convenient path. Indeed, the above computation on cospans gives us an isomorphism of fiber functors
$$\sigma:F_{\bar{y}} \to F_{\sigma(\bar{y})}$$ and the map on automorphism groups is just given by conjugation by this isomorphism. This means that the map $\pi_1(Y,\sigma(\bar{y})) \to \pi_1(Y,\bar{y})$ is, up to conjugation, given as the inverse of the map $\pi_1(Y,\bar{y}) \to \pi_1(Y,\sigma(\bar{y}))$.
This shows that the composite $\pi_1(Y,\bar{y}) \to \pi_1(Y,\sigma(\bar{y})) \to \pi_1(Y,\bar{y})$ is given by conjugation.
