[proofplan] We first verify that each conjugation map $\iota_a$ is an automorphism of $G$, so that $\Phi$ is a well-defined function into $\operatorname{Aut}(G)$. Then we compare $\Phi(ab)$ with the composition $\Phi(a)\circ\Phi(b)$ by evaluating both maps on an arbitrary element of $G$. The identity $(ab)^{-1}=b^{-1}a^{-1}$ gives exactly the compatibility required for a group homomorphism. [/proofplan] [step:Show each conjugation map is an automorphism of $G$] Let $a \in G$ be fixed, and define \begin{align*} \iota_a:G&\to G\\ x&\mapsto axa^{-1}. \end{align*} We show that $\iota_a$ is a group automorphism. For arbitrary $x,y \in G$, associativity in $G$ gives \begin{align*} \iota_a(xy) &= a(xy)a^{-1} \\ &= ax(e)y a^{-1} \\ &= ax(a^{-1}a)y a^{-1} \\ &= (axa^{-1})(aya^{-1}) \\ &= \iota_a(x)\iota_a(y), \end{align*} where $e \in G$ denotes the identity element. Thus $\iota_a$ is a group homomorphism. Define the map \begin{align*} \iota_{a^{-1}}:G&\to G\\ x&\mapsto a^{-1}xa. \end{align*} For every $x \in G$, \begin{align*} (\iota_a\circ\iota_{a^{-1}})(x) &= \iota_a(a^{-1}xa) \\ &= a(a^{-1}xa)a^{-1} \\ &= x, \end{align*} and similarly \begin{align*} (\iota_{a^{-1}}\circ\iota_a)(x) &= \iota_{a^{-1}}(axa^{-1}) \\ &= a^{-1}(axa^{-1})a \\ &= x. \end{align*} Hence $\iota_{a^{-1}}$ is the inverse function of $\iota_a$, so $\iota_a$ is bijective. Since $\iota_a$ is both a group homomorphism and bijective, $\iota_a \in \operatorname{Aut}(G)$. [guided] Fix an element $a \in G$. The first point is that the formula $x \mapsto axa^{-1}$ must actually define an automorphism of $G$, because the codomain of $\Phi$ is $\operatorname{Aut}(G)$. Define \begin{align*} \iota_a:G&\to G\\ x&\mapsto axa^{-1}. \end{align*} This is a function from $G$ to $G$ because $G$ is closed under multiplication and inverses. We verify first that $\iota_a$ preserves multiplication. Let $x,y \in G$ be arbitrary. Using associativity, the identity relation $e=a^{-1}a$, and the definition of $\iota_a$, we compute \begin{align*} \iota_a(xy) &= a(xy)a^{-1} \\ &= ax(e)y a^{-1} \\ &= ax(a^{-1}a)y a^{-1} \\ &= (axa^{-1})(aya^{-1}) \\ &= \iota_a(x)\iota_a(y). \end{align*} Thus $\iota_a$ is a group homomorphism $G \to G$. To prove that this homomorphism is an automorphism, we exhibit its inverse explicitly. Define \begin{align*} \iota_{a^{-1}}:G&\to G\\ x&\mapsto a^{-1}xa. \end{align*} For every $x \in G$, composition with $\iota_a$ gives \begin{align*} (\iota_a\circ\iota_{a^{-1}})(x) &= \iota_a(a^{-1}xa) \\ &= a(a^{-1}xa)a^{-1} \\ &= (aa^{-1})x(aa^{-1}) \\ &= exe \\ &= x. \end{align*} The other composition is also the identity map on $G$: \begin{align*} (\iota_{a^{-1}}\circ\iota_a)(x) &= \iota_{a^{-1}}(axa^{-1}) \\ &= a^{-1}(axa^{-1})a \\ &= (a^{-1}a)x(a^{-1}a) \\ &= exe \\ &= x. \end{align*} Therefore $\iota_{a^{-1}}$ is the inverse function of $\iota_a$. Since $\iota_a$ is a bijective group homomorphism from $G$ to itself, it is an automorphism of $G$. Hence $\iota_a \in \operatorname{Aut}(G)$. [/guided] [/step] [step:Define the map into the automorphism group] By the previous step, for every $a \in G$ the map $\iota_a$ belongs to $\operatorname{Aut}(G)$. Therefore the assignment \begin{align*} \Phi:G&\to \operatorname{Aut}(G)\\ a&\mapsto \iota_a \end{align*} is a well-defined function. [/step] [step:Verify preservation of the group operation] Let $a,b \in G$ be arbitrary. The group operation on $\operatorname{Aut}(G)$ is composition of automorphisms, so we must prove \begin{align*} \Phi(ab)=\Phi(a)\circ\Phi(b). \end{align*} Both sides are functions $G \to G$. Let $x \in G$ be arbitrary. Using the definition of $\Phi$, the definition of $\iota_a$, associativity in $G$, and the inverse identity $(ab)^{-1}=b^{-1}a^{-1}$, we compute \begin{align*} \Phi(ab)(x) &= \iota_{ab}(x) \\ &= (ab)x(ab)^{-1} \\ &= (ab)x(b^{-1}a^{-1}) \\ &= a(bxb^{-1})a^{-1} \\ &= \iota_a(\iota_b(x)) \\ &= (\Phi(a)\circ\Phi(b))(x). \end{align*} Since the equality holds for every $x \in G$, the two functions are equal: \begin{align*} \Phi(ab)=\Phi(a)\circ\Phi(b). \end{align*} Thus $\Phi$ preserves the group operation. [/step] [step:Conclude that $\Phi$ is a group homomorphism] The domain $G$ is a group, the codomain $\operatorname{Aut}(G)$ is a group under composition, and the previous step proves that for all $a,b \in G$, \begin{align*} \Phi(ab)=\Phi(a)\Phi(b), \end{align*} where multiplication in $\operatorname{Aut}(G)$ means composition. By the definition of a [group homomorphism](/page/Group%20Homomorphism), $\Phi:G\to\operatorname{Aut}(G)$ is a group homomorphism. [/step]