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So you're giving an endomorphism $\varphi:F_n\to F_n$, such that $\varphi(x_i)=a_i^{-1}x_ia_i$, and you want to know if this is an isomorphism?

There is a nice algorithm due to Stallings, called Stallings foldings, which will quickly tell you the answer. One may assume that $a_i$ is a reduced word and does not begin in $x_i$, x_i^{-1}$, and that the$a_i$'s share no common postfix. If you take a wedge of loops as a$K(F_n,1)$, with a loop for each generator$x_i$, then the map$\varphi$may be realized by a map between wedges of loops, where the loop$x_i$goes to a loop represented by$a_i^{-1}x_ia_i$. The domain graph gets an induced cell decomposition, where the loop$x_i$gets subdivided into$|a_i^{-1}x_ia_i|$edges. Then you can "fold" the domain graph, making it shorter by identifying edges stemming from a vertex which map to the same edge in the target. If you iterate until no folds are available, then the map will be an isomorphism if and only if the map is a homeomorphism at the end. From this, one gets a simple sufficient criterion for the endomorphism to not be an isomorphism, namely there are no folds available. This happens if the elements$a_i$end in different generators and their inverses as reduced words. In general, though, I think one would have to carry out Stallings algorithm to get a necessary criterion. 1 So you're giving an endomorphism$\varphi:F_n\to F_n$, such that$\varphi(x_i)=a_i^{-1}x_ia_i$, and you want to know if this is an isomorphism? There is a nice algorithm due to Stallings, called Stallings foldings, which will quickly tell you the answer. One may assume that$a_i$is a reduced word and does not begin in$x_i$, and that the$a_i$'s share no common postfix. If you take a wedge of loops as a$K(F_n,1)$, with a loop for each generator$x_i$, then the map$\varphi$may be realized by a map between wedges of loops, where the loop$x_i$goes to a loop represented by$a_i^{-1}x_ia_i$. The domain graph gets an induced cell decomposition, where the loop$x_i$gets subdivided into$|a_i^{-1}x_ia_i|$edges. Then you can "fold" the domain graph, making it shorter by identifying edges stemming from a vertex which map to the same edge in the target. If you iterate until no folds are available, then the map will be an isomorphism if and only if the map is a homeomorphism at the end. From this, one gets a simple sufficient criterion for the endomorphism to be an isomorphism, namely there are no folds available. This happens if the elements$a_i\$ end in different generators and their inverses as reduced words. In general, though, I think one would have to carry out Stallings algorithm to get a necessary criterion.