Exhibit the correspondence in the Correspondence Theorem. Here is the whole problem.  I answered the first two parts, but I can't get down the third part.
Problem
Consider the group $D_{4} = \langle x,y:x^{2}=1, y^{4}=1, yx=xy^{3}\rangle$ and the homomorphism $\Phi : D_{4} \rightarrow Aut(D_{4})$ defined by $\Phi (g) = \phi _{g}$.  Define $\phi : G \rightarrow G$ by $\phi _{g}(x)=g^{-1}xg$.
There are three parts for the whole problem, which are:
(a) Determine $K = ker(\Phi)$
(b) Write down the cosets of K.
(c) Let $Inn(D_{4}) = \Phi(D_{4})$.  Then, $\Phi : D_{4} \rightarrow Inn(D_{4})$ is surjective.  Exhibit the correspondence in the Correspondence Theorem explicitly.
My Attempt
✓ I've found the kernel, which is $ker(\Phi) = \{e,y^{2}\}$
✓ There are four distinct cosets of K, which I found.  I don't need help with this.
✗ I haven't completely exhibit the correspondence of subgroups. Here is what I have:


*

*$ker(\Phi) \leftrightarrow \{e\}$

*$D_{4} \leftrightarrow Inn(D_{4})$


I know that $|D_{4}|=8$ and since $\Phi$ is surjective and $|ker(\Phi)|=2$, $|Inn(D_{4})|=4$.  If I'm right, there should be 4 pairs of corresponding subgroups.  I only got down two.
Any advices or comments? Or probably some hints?
 A: Note that always $\,\ker\phi=Z(G)\,$ , so in fact you have that
$$G/Z(G)\cong Inn(G)$$
and since $\,G/Z(G)\,$ can never be cyclic non-trivial, it must be that $\,Inn(D_4)\cong C_2\times C_2\,$ . This last group has three non-trivial proper subgroups (all of which are cyclic of order two), not four, so I'm not sure why do you expect four...and they correspond to the cosets $\,x\,,\,y\,,\,xy\,$ ...
Added: As already said, $\,Z:=Z(D_4)=\{1,y^2\}\,$ , but then note that $\,xZ=xy^2Z\,$ , and in fact $\,D_4/Z=\{\;\bar 1=Z\,,\,xZ\,,\,yZ\,,\,xyZ\,\;\}\,$ , so if we denote by $\,f:D_4\to D_4/Z\,$ the bijective correspondence between the subgps. of the quotient $\,G/Z\,$ and the subgps. of $\,D_4\;$ containing $\,Z\,$ we get :
$$\begin{align*}(1)&Z&\longleftrightarrow &\bar 1=Z\\
(2)&\{1,x,y^2,xy^2\}&\longleftrightarrow &\langle\; xZ\;\rangle\\
(3)&\{1,y,\,y^2,y^3\}&\longleftrightarrow &\langle\; yZ\;\rangle\\
(4)&\{1,xy,\,xy^3,y^2\}&\longleftrightarrow &\langle\; xyZ\;\rangle\end{align*}$$
A: Its important to properly first understand $\Phi$. Now as you say, $Inn(D_4)$ has four functions in it. If you have done the math you may have already seen that $\Phi$ is as given below:
$\Phi(1)=\Phi(y^2)=Id$
$\Phi(x)=\Phi(xy^2)=\phi_x$
$\Phi(y)=\Phi(y^3)=\phi_y$
$\Phi(xy)=\Phi(xy^3)=\phi_{xy}$
Now we have $D_4/ker(\Phi)\approx Inn(D_4)$. What is this isomorphism explicitly? By the proof of the fundamental theorem of homomorphism we know that this isomorphism is precisely:
$K\to Id$
$Kx\to \phi_x$
$Ky\to \phi_y$
$Kxy\to\phi_{xy} $
where $K,Kx,Ky,Kxy$ are the four distinct cosets already obtained in a previous part.
Now, there are two groups of order 4: the cyclic group $\mathbb{Z}_4$ and the non cyclic Klein's 4 group $\mathbb{Z}_2\times \mathbb{Z}_2$. Since $Inn(D_4)$ is not cyclic (if it were then $D_4$ would be abelian, confer this post and Don Antonio's comment) and so has the following subgroups: $\{Id\},\{Id,\phi_x\},\{Id,\phi_y\},\{Id,\phi_{xy}\}$. This follows because every non cyclic group of order 4 has a structure similar to the Klein's group.
To find the correspondence it suffices to look at the images of each of these under the isomorphism. Hence the corresponding subgroups are:
$\{K\}\leftrightarrow \{Id\}$ 
$\{K,Kx\}\leftrightarrow\{Id,\phi_x\}$
$\{K,Ky\}\leftrightarrow\{Id,\phi_y\}$
$\{K,Kxy\}\leftrightarrow\{Id,\phi_{xy}\}$ 
This completes the proof.
A: Hint: The remaining subgroups will be cyclic subgroups of order $2$. There will be  $3$ of them (since $Inn(G)$ is not cyclic for $G$ nonabelian).
