This is unusual terminology, but legitimate. We usually talk about a relation being transitive, meaning that $aRb,bRc$ implies $aRc$. If we take $R$ to be the relation which has $aRf(a)$, then it will be transitive if $aRf(a)$ and $f(a)Rb$ implies $aRb$. But since $f()$ is a function that means $b$ must equal $f(a)$. In other words, the point $f(a)$ which belongs to the image of $f()$ is fixed.
That explains the terminology. Answering the question requires careful counting. Let us count functions with 1,2,3,4 fixed points. Note that the image has at least one point and that is fixed, so those are the only possibilities.
Taking the easiest first, suppose it has 4 fixed points. There is only one such function, it has $f(a)=a,f(b)=b,f(c)=c,f(d)=d$.
Suppose there is one fixed point. That means the image has only one point. So there are just 4 such functions. For example, $f(a)=f(b)=f(c)=f(d)=a$.
Now suppose there are three fixed points. Suppose they are $a,b,c$. That gives us $f(a),f(b),f(c)$. There are now three choices for $f(d)$, it can be either $a,b$ or $c$. So that gives us a total of $4\times 3=12$ functions with 3 fixed points.
Finally, consider two fixed points. We can choose the 2 in 6 ways. Suppose they are $a,b$. Now there are two choices for each of $f(c),f(d)$ - they can each be $a$ or $b$. So that is 24 functions with 2 fixed points.
Adding $\large 4+ 1 + 12 + 24=41$.