If $f^n$ is the identity map then prove that $f$ is bijection
Note that $f^{n-1}$ is the inverse of $f$. It follows that $f$ is a bijection.
I would avoid a proof by contradiction here. In that case, your proof of injectivity becomes:
"Suppose $f(x_1)=f(x_2)$. Then $x_1=f^{n-1}(f(x_1))=f^{n-1}(f(x_2))=x_2$."
For surjectivity: $f^n=\mathrm{id}_A$, so $f(f^{n-1}(y))=y$, meaning $f(x)=y$ where $x=f^{n-1}(y)$.
Recall:
Theorem. Let $F \colon A \to B$ and $G \colon B \to C$.
- If $G \circ F$ is injective, then $F$ is injective.
- If $G \circ F$ is surjective, then $G$ is surjective.
Use that theorem for $f \colon A \to A$.
- Since $\operatorname{id}_A = f^n = f^{n-1} \circ \color{red}f$ is injective, $f$ is injective.
- Since $\operatorname{id}_A = f^n = \color{red}f \circ f^{n-1}$ is surjective, $f$ is surjective.
Therefore, $f$ is bijective.