Suppose we consider the dicyclic group of order 12 given by the following relation $$\mathrm{Dic}_{12} = \langle x,y : x^3 = y^4 = e, yxy^{-1}x = e \rangle.$$
Consider the following subgroups of $\mathrm{Aut}(G)$:
$$\mathrm{Stab}(x) = \{ \phi \in\mathrm{Aut}(G) : \phi(x) = x\}$$
$$\mathrm{Stab}(y) = \{ \phi \in\mathrm{Aut}(G) : \phi(y) = y\}.$$
I want to prove that $\mathrm{Aut}(G) \simeq\mathrm{Stab}(x) \rtimes\mathrm{Stab}(y) \simeq D_{12}$.
The idea that I am having for attacking this problem is first proving that $|Aut(G)| = 12$ using combinatorics then also calculating $|Stab(x)|$ and $|Stab(y)|$ using combinatorics and proving that $Stab(x)$ is normal and using that will give us the semidirect isomorphism, but I am not sure how to do this in detailed fashion and proving that $Aut(G) \simeq D_{12}$.
Here is a sketch solution. You can fill in the details yourself.
Note first that $\langle x \rangle$ is characteristic in $G$. So automorphisms of $G$ map $x$ to $x$ or to $x^2$. Since the automorphism $c_y$ (conjugation by $y$) has order $2$ and lies in ${\rm Stab}(y)$ but not in ${\rm Stab}(x)$, we have ${\rm Aut}(G) = {\rm Stab}(x) \rtimes {\rm Stab}(y)$.
Now an automorphism in ${\rm Stab}(x)$ is determined by its action on $y$, and there are exactly $6$ elements of order $4$ in $G$, namely $y^{\pm 1}x^{k}$ with $k = 0,1,2$, so $|{\rm Stab}(x)| \le 6$. On the other hand, the automorphism $\alpha$ of ${\rm Stab}(x)$ with $\alpha:x \mapsto x$, $\alpha:y \mapsto y^{-1}x$ has order $6$.
To prove that ${\rm Aut}(G) \cong D_{12}$, check that $c_y\alpha$ has order $2$.