$\psi: G \to \text{Inn}(G)$ where $\psi(x)=\varphi_x$. How do I show $\psi$ is well defined? Inn$(G)=\{\varphi_g \in \text{Aut}(G) \mid g \in G\}$
If $\varphi_g, \varphi_h \in \text{Inn}(G)$, then
$$\varphi_g \varphi_h (x) =\varphi_g(hxh^{-1})=ghxh^{-1}g^{-1}=ghx(gh)^{-1}=\varphi_{gh} \in \text{Inn}(G)$$
Also, since $\varphi_g\varphi_g^{-1}=x$, and $\varphi_g\varphi_{g^{-1}}=x$,
$$\varphi_g^{-1}=\varphi_{g^{-1}} \in \text{Inn}(G)$$
So, Inn$(G) \le \text{Aut}(G)$.
Define $\psi: G \to \text{Inn}(G)$ where $\psi(x)=\varphi_x$
How do I show $\psi$ is well defined?
I need to show that if $a=b$, then $\psi(a)=\psi(b)$
 A: It's well-defined because you defined it.
The times when you need to check something is well-defined is when you give a "definition" that involves making some extra choices; you then have to check that your construction is independent of these choices. You didn't make any choices so your question doesn't seem to have any content.
A: Well-definedness is not needed to be shown as there is no choice involved: Given $x\in G$ you are to exhibit  some $\psi(x)\in\operatorname{Inn}(G)$. Since you are given $x\in G$, no-one can prevent you from considering the inner automorphism $\phi_x$.

Just to make clear where we do need to show well-definedness: Suppose we wanted to define a map in the reverse direction, i.e., $\Psi\colon \operatorname{Inn}(G)\to G$, $\phi_x\mapsto x$. In this case, we'd have to show that $x$ can be determined from the given inner automorphism. That is, if $f\in\operatorname{Inn}(G)$, then there exists only one possible choice $x\in G$ such that $f=\phi_x$. This is in general not the case, hence the reverse map is not well-defined.
