Central automorphisms of a group form a subgroup of the automorphism group I read this question about central automorphisms. The OP states that he could easily proof that the set of central automorphisms forms a subgroup of the automorphism group, however, I am not able to do this. 
A central automorphism is a automorphism $\omega$ such that for every $g \in G$ the element $g^{-1}\omega\left(g\right) \in Z(G)$, where $Z(G)$ is the center of the group $G$. 
First of all, the identity map is a central automorphism, so the set of central automorphisms is not empty. 
Let $\omega, \theta$ be central automorphisms, then I am trying to show that $\omega \circ \theta^{-1}$ is such an automorphism itself (the subgroup criterion then assures that this set is a subgroup of the automorphis group). Let $x,g \in G$ be two arbitrary elements, then I need to show that 
$$g^{-1} \omega(\theta^{-1}(g))x = xg^{-1}\omega(\theta^{-1}(g))$$
However, I do not see how to use the information that $\omega, \theta$ are central homomorphisms... 
Any hints on how to finish/remarks on my approach are highly appreciated. 
 A: It's easier to show, separately, that $\omega\circ\theta$ and $\omega^{-1}$ are central automorphisms.
Since $\omega$ is a central automorphism, we have
$$ (\omega\circ\theta)(g) =\omega(\theta(g))\in\theta(g)Z(G).$$
Since $\theta$ is a central automorphism, $\theta(g)\in gZ(G)$, so in fact $(\omega\circ\theta)(g)\in gZ(G)$. Thus, $\omega\circ\theta$ is a central automorphism.
To show that $\omega^{-1}$ is a central automorphism, let $g\in G$, so that $\omega(g)\in gZ(G)$.  Then $g\in \omega^{-1}(gZ(G)) = \omega^{-1}(g)Z(G)$, as $Z(G)$ is characteristic. So, $g = \omega^{-1}(g)z$, for $z\in Z(G)$, and hence $\omega^{-1}(g) = gz^{-1}\in gZ(G)$. This shows that $\omega^{-1}$ is a central automorphism.
A: First an abstract proof:
Surjective homomorphisms of groups stabilize the center, therefore we have a natural homomorphism
$$\text{Aut}(G) \to \text{Aut}(G/Z(G))$$
whose kernel is precisely the central automorphisms, as for $\phi \in \text{Aut}(G)$ then $[\phi(g)] = [g]$ in $G/Z(G)$ is equivalent to the difference $\phi(g)g^{-1}$ being in $Z(G)$. $\square$

Second, it is quite possible to use the one-step subgroup test as in the OP's attempt to your advantage here - just make a substitution $g=\theta(h)$:
$$
\begin{aligned}
g^{-1}.(\omega \circ \theta^{-1})(g)
&= g^{-1}.\omega(\theta^{-1}(g))\\
&= \theta(h)^{-1}.\omega(h)\\
&= \theta(h^{-1}).\omega(h)\\
&= h.\theta(h^{-1}).h^{-1}.\omega(h)
\end{aligned}
$$
$\square$
This proof is also nice because each line is an assumption - substitution is bijectivity, next step is homomorphism, last step is centrality.
