There are four elements in the Klein -4 group, with three elements non-trivial.
Order of all non-trivial elements is $2,$ and product of any two such elements is another.
The group table is: \begin{array}{|c|c|c|} \hline &e&x&y&xy\\ \hline e&e&x&y&xy \\ \hline x&x&e&xy&y \\ \hline y&y&xy&e&x \\ \hline xy&xy&y&x&e \\ \hline \end{array}
For having endomorphism, need for any two elements $a,b\in K_4,$ need $\phi(a\times b)= \phi(a)\times \phi(b).$
The possible number of values of maps for pairs: $\phi(a), \phi(b),$ need be found for satisfying the homomorphism property.
First consider the trivial map of all elements mapping to $e.$
0. $\phi(e)=
\phi(x)= \phi(y)= \phi(xy)= e.$
Next, consider non-trivial homomorphisms, with $\times$ being the group operation of composition.
$\phi(e)=e, \phi(x)=x, \phi(y)=y, \phi(xy)= xy.$
Get, $\phi(x\times y)= \phi(xy)= xy= \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=x, \phi(y)=x, \phi(xy)= e.$
Get, $\phi(x\times y)= \phi(xy)= e= \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=x,\phi(y)=xy,\phi(xy)= y.$
Get, $\phi(x\times y)= \phi(xy)= y = \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=x, \phi(y)=e, \phi(xy)= x.$
Get, $\phi(x\times y)= \phi(xy)= x = \phi(x) \times \phi(y).$
$\phi(e)=e, \phi(x)=y, \phi(y)=x, \phi(xy)= xy.$
Get, $\phi(x\times y)= \phi(xy)= xy = yx = \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=y,\phi(y)=y, \phi(xy)= e.$
Get, $\phi(x\times y)= \phi(xy)= e = \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=y,\phi(y)=xy,\phi(xy)= x.$
Get, $\phi(x\times y)= \phi(xy)= x = \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=y, \phi(y)=e, \phi(xy)= y.$
Get, $\phi(x\times y)= \phi(xy)= y = \phi(x) \times \phi(y).$
$\phi(e)=e, \phi(x)=xy, \phi(y)=y, \phi(xy)= x.$
Get, $\phi(x\times y)= \phi(xy)= x= \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=xy, \phi(y)=x, \phi(xy)= y.$
Get, $\phi(x\times y)= \phi(xy)= y = \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=xy,\phi(y)=xy,\phi(xy)= e.$
Get, $\phi(x\times y)= \phi(xy)= e = \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=xy, \phi(y)=e, \phi(xy)= xy.$
Get, $\phi(x\times y)= \phi(xy)= xy = \phi(x) \times \phi(y).$
$\phi(e)=e, \phi(x)=e, \phi(y)=x, \phi(xy)= x.$
Get, $\phi(x\times y)= \phi(xy)= x= \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=e, \phi(y)=y, \phi(xy)= y.$
Get, $\phi(x\times y)= \phi(xy)= y = \phi(x) \times \phi(y).$$\phi(e)=e, \phi(x)=e,\phi(y)=xy,\phi(xy)= xy.$
Get, $\phi(x\times y)= \phi(xy)= xy = \phi(x) \times \phi(y).$
So, there are $16$ endomorphisms, as both $\phi(a), \phi(b)$have four choices each. Hence, $4\star4=16.$
Kindly vet the above.
