This is a simple computation (write down your matrix, and see what it means for a two-by-two matrix to commute with it, or to normalize the subgroup). When doing the computation, it is useful to remember that two matrices commute if (and only if) they have the same eigenvectors.
EDIT In view of the fancy-pants arguments in the other answers, I do the computation for a parabolic in your notation $(1\ x)(0\ 1).$ Conjugating by $(a,\ b)(c,l\ d)$ gives you
$(1 - a c x,\ a^2 x)(-c^2 x, \ 1+ a c x).$ from the lower left corner we see that $c=0.$ That means (since we are in $SL(2, Z),$ that would not work in $SL(2, R)$) that either $a=d=1$ or $a=d=-1,$ so the normalizer in $PSL(2, Z)$ is just the subgroup itself, in $SL$ you throw in the center. The computation for elliptic (or hyperbolic) element is equally difficult.