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No, consider the map $F_3 \to F_3$ given by $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, x_3[x_2, x_1]], x_3 \mapsto x_3[x_2, x_1], . $$ It is an IA automorphism: it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x_1, x_3[x_2, x_1], x_2 [x_1, x_3[x_2, x_1]] \}$ generates and $F_3$ is Hopfian.

Further, if you kill $x_3$ (that is, $n = 3$ in the OP's question), then the resulting map under consideration in the question is: $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, [x_2, x_1]], $$ which is clearly not an isomorphism (the word $x_2 [x_1, [x_2, x_1]] = x_2x_1^{-1} (x_2^{-1} x_1^{-1} x_2 x_1) x_1 (x_1^{-1} x_2^{-1} x_1 x_2)$ in reduced form begins and ends with $x_2$).

No, consider the map $F_3 \to F_3$ given by $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, x_3[x_2, x_1]], x_3 \mapsto x_3[x_2, x_1]. $$ It is an IA automorphism: it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x_1, x_3[x_2, x_1], x_2 [x_1, x_3[x_2, x_1]] \}$ generates and $F_3$ is Hopfian.

Further, if you kill $x_3$ (that is, $n = 3$ in the OP's question), then the resulting map under consideration in the question is: $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, [x_2, x_1]], $$ which is clearly not an isomorphism (the word $x_2 [x_1, [x_2, x_1]] = x_2x_1^{-1} (x_2^{-1} x_1^{-1} x_2 x_1) x_1 (x_1^{-1} x_2^{-1} x_1 x_2)$ in reduced form begins and ends with $x_2$).

No, consider the map $F_3 \to F_3$ given by $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, x_3[x_2, x_1]], x_3 \mapsto x_3[x_2, x_1], . $$ It is an IA automorphism: it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x_1, x_3[x_2, x_1], x_2 [x_1, x_3[x_2, x_1]] \}$ generates and $F_3$ is Hopfian.

Further, if you kill $x_3$, the resulting map under consideration in the question is: $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, [x_2, x_1]], $$ which is clearly not an isomorphism (the word $x_2 [x_1, [x_2, x_1]] = x_2x_1^{-1} (x_2^{-1} x_1^{-1} x_2 x_1) x_1 (x_1^{-1} x_2^{-1} x_1 x_2)$ in reduced form begins and ends with $x_2$).

No, consider the map $F_3 \to F_3$ given by $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, x_3[x_2, x_1]], x_3 \mapsto x_3[x_2, x_1], . $$ It is an IA automorphism: it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x_1, x_3[x_2, x_1], x_2 [x_1, x_3[x_2, x_1]] \}$ generates and $F_3$ is Hopfian.

Further, if you kill $x_3$ (that is, $n = 3$ in the OP's question), then the resulting map under consideration in the question is: $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, [x_2, x_1]], $$ which is clearly not an isomorphism (the word $x_2 [x_1, [x_2, x_1]] = x_2x_1^{-1} (x_2^{-1} x_1^{-1} x_2 x_1) x_1 (x_1^{-1} x_2^{-1} x_1 x_2)$ in reduced form begins and ends with $x_2$).

No, consider the map $F_3 \to F_3$ given by $$ x_1 \mapsto x_1, x_2 \mapsto x_2[x_3, x_1], x_3 \mapsto x_3 [x_1, x_2[x_3, x_1]]. $$ It is an IA automorphism: it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x_1, x_2[x_3, x_1], x_3 [x_1, x_2[x_3, x_1]] \}$ generates and $F_3$ is Hopfian.

Further, if you kill $x_2$, you get the map $$ x_1 \mapsto x_1, x_3 \mapsto x_3 [x_1, [x_3, x_1]], $$ which is clearly not an isomorphism (the word $x_3 [x_1, [x_3, x_1]] = x_3 x_1^{-1} (x_3^{-1} x_1^{-1} x_3 x_1) x_1 (x_1^{-1} x_3^{-1} x_1 x_3)$ in reduced form begins and ends with $x_3$).

No, consider the map $F_3 \to F_3$ given by $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, x_3[x_2, x_1]], x_3 \mapsto x_3[x_2, x_1], . $$ It is an IA automorphism: it clearly induces the identity map on abelianization. Further, it is an isomorphism because the set $\{ x_1, x_3[x_2, x_1], x_2 [x_1, x_3[x_2, x_1]] \}$ generates and $F_3$ is Hopfian.

Further, if you kill $x_3$, the resulting map under consideration in the question is: $$ x_1 \mapsto x_1, x_2 \mapsto x_2 [x_1, [x_2, x_1]], $$ which is clearly not an isomorphism (the word $x_2 [x_1, [x_2, x_1]] = x_2x_1^{-1} (x_2^{-1} x_1^{-1} x_2 x_1) x_1 (x_1^{-1} x_2^{-1} x_1 x_2)$ in reduced form begins and ends with $x_2$).