Answer: B

Solution:

Reflecting the graph of $y=f(x)$ over $y=x$ produces the graph of its inverse, so the graph is symmetric about $y=x$ exactly when $f$ equals its own inverse. For $f(x)=\tfrac{ax+b}{cx+d}$ this happens in two ways: when $a+d=0$ with $ad-bc\ne0$ (a genuine involution, including the slope$-1$ lines when $c=0$), or when $f$ is the identity $y=x$ ($b=c=0,\ a=d\ne0$).

For $a+d=0,$ set $d=-a;$ the determinant $-a^2-bc$ must be nonzero, so we need $a^2+bc\ne0,$ together with $(c,d)\ne(0,0).$ When $a=0,$ both $b$ and $c$ must be nonzero, giving $10^2=100$ choices. For each nonzero $a,$ start with $11^2=121$ choices of $(b,c).$ If $|a|=1,3,4,$ or $5,$ exactly $2$ pairs satisfy $bc=-a^2;$ if $|a|=2,$ exactly $6$ pairs do. Thus the genuine involutions number $$\begin{gathered} 100+8(121-2)+2(121-6)\\ {}=1282. \end{gathered}$$ The identity case adds $10$ more ($a=d\in\{\pm1,\ldots,\pm5\}$), for a total of $1282+10=1292.$

Thus, the correct answer is B.