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For a prime $p>2$, a diagonal matrix with two distinct nonzero eigenvalues exists. The same common-eigenspace argument makes the normalizer of a diagonal subgroup of GL2 precisely the monomial matrices. There are $(p-1{)}^2$ choices of nonzero entries for each of two permutation patterns, givingAt $p=2$, however, the invertible diagonal subgroup is trivial. Every invertible matrix normalizes it, soThe first factor chooses a nonzero first column and the second chooses a column outside its one-dimensional span. The exceptional answer is necessary because the printed question permits every prime.

The coordinate axes are exactly the simultaneous eigenspaces of all invertible diagonal matrices in $T$. A matrix normalizing $T$ must permute these common eigenspaces: if $tv=\chi (t)v$, then $t(gv)=g(g^{-1}tg)v$ is again a scalar multiple of $gv$. Thus a normalizer element has exactly one nonzero entry in each row and column, a monomial matrix. In dimension two it is either diagonal or diagonal times $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Conversely $J$ swaps the diagonal entries and normalizes $T$. ThereforeThis is the normalizer of a diagonal subgroup of GL2; the quotient records the permutation of the two axes.