Let $A$ be a group and fix $b ∈ A$. Show that the map $ϕ : A → A$ given by $ϕ(a) := b^{-1}ab$ is an automorphism of A. Let $A$ be a group and fix $b ∈ A$. Show that the map $ϕ : A → A$ given by $ϕ(a) := b^{-1}ab$ is an automorphism of A.
Im not even sure how to start this as I am struggling with understanding what automorphism is. It's an isomorphism to itself, so i need to find a way to show this??
 A: The definition of an isomorphism is: a morphism $f:X\to Y$ is an isomorphism if there exists a morphism $g:Y\to X$ such that $f\circ g=id_{Y}$ and $g\circ f=id_X$. I use the term "morphism" because it is a more general notion but in your case it is simply a group homomorphism.
An automorphism is an isomorphism $f:X\to X$.
So first you need to show that $\phi$ is a group homomorphism. This follows from
$$\phi(xy)=b^{-1}xyb=b^{-1}x(bb^{-1})yb=(b^{-1}xb)(b^{-1}yb)=\phi(x)\phi(y)$$
The second requirement follows simply from noticing that the inverse of $\phi$ is $\psi$ given by $\psi(x)=bxb^{-1}$. Note that the inverse is automatically a group homomorphism because the previous statement was proved for an arbitrary $b\in G$.

Side note: While in the context of groups an isomorphism is a bijective homomorphism this is not the case in other places. For example in the context of topological spaces the continuous bijective map need not be a homeomorphism (equivalent of isomorphism for continuous maps). So obviously you can use the "bijective homomorphism" definition in the context of groups but just be aware that there's a subtle difference between a morphism being "bijective" and "invertible".
