Conjugate relations? I was going through time reversal symmetry and came across the term 'conjugate relation'. The text is

Next, in classical mechanics, the initial conditions of a motion $x(t)$ transform under time reversal according to $(x_0, p_0) → (x_0,−p_0)$, so we postulate that the time-reversal operator $\theta$ in quantum
  mechanics should satisfy the conjugation relations,
$\theta x \theta^\dagger = x$,
$\theta p \theta^\dagger = −p$.  

My question is what meant by conjugate relation? 
As far as I know conjugate relation is related with the conjugate momenta and position in classical mechanics. How it is related here and how can one convey that time reversal operator cannot be unitary by looking at $\theta x \theta^\dagger = x$, $\theta p \theta^\dagger = −p$.)
 A: the operation $X\mapsto A^{-1}XA$ is usually called conjugation of $X$ by $A$ because this is the language used in group theory. I assume that by $\dagger$ the author means inverse, because antilinear/anti-unitary  maps like time reversal do not have an adjoint which is what the $\dagger$ usually denotes.
The "group theory" mention is not important. It's just the origin of the name.  When we diagonalize a matrix by 
$$
X\mapsto A^{-1}XA= {\rm diag}(\lambda_1,\lambda_2\ldots )
$$
the operation is a called a similarity transformation or a 
conjugation and $X$ and the diagonal matrix are said to be similar or conjugate.  
Conjugation  is to be distingushed from a diagonalizing matrix by
$$
X\mapsto A^TXA= {\rm diag}(\lambda_1,\lambda_2\ldots )
$$
as we do for quadratic forms such as the kinetic energy in a small vibrations problem. In this case $X$ and the diagonal matric are said to be congruent rather than conjugate.
Time reversal has to be antilinear in order to deal with the "$i$" in the time dependent Schrodinger equation. If $\psi(x,t)$ satisfies it and the potential is real, then so does $\psi^*(x,-t)$. So the wavefunction has to be be complex-conjugated and this is an antilinear process:
$$
\theta( \alpha |{\bf a}\rangle +\beta |{\bf b}\rangle)= \alpha^* (\theta |{\bf a}\rangle) +\beta^* (\theta|{\bf b}\rangle )
$$
Antilinear/antiuitary  maps are tricky and confusing things. 
An operator $\Omega$ is said to be antiunitary with respect to the usual quantum mechanics  (conjugate-symmetric and sesquilinear) inner product $\langle{\phantom -},{\phantom-}\rangle$ if
$$
\langle{\Omega {\bf a}}, {\Omega {\bf b}}\rangle  =\langle{{\bf a}},{{\bf b}}\rangle ^*= \langle{\bf b},{\bf a}\rangle. 
$$
 Consider the vector
$$
{\bf X}= \Omega( \alpha {\bf a}+\beta {\bf b})- \alpha^* (\Omega {\bf a}) -\beta^* (\Omega{\bf b}).
$$
Using   the definition of antiunitarity and the antilinearity of $\langle{\phantom -},{\phantom-}\rangle $ in its  first slot and linearity in the second,  we can expand out   $\|{\bf X}\|^2=\langle{{\bf X}},{{\bf X}}\rangle$ and find that it is zero. For a  positive definite  inner product a vanishing norm  implies  that ${\bf X}=0$, and so for such a product we have 
$$
\Omega( \alpha {\bf a}+\beta {\bf b})= \alpha^* (\Omega {\bf a}) +\beta^* (\Omega{\bf b}).
$$
Thus an  antiunitary operator acting on a positive-definite Hilbert space is necessarily antilinear. 
One  consequence of the antilinearity  is that there is no obvious way to define an  adjoint $\Omega^\dagger$.   The standard definition of the adjoint (i.e the "Hermitian conjugate")  $\Omega^\dagger$ with respect to an inner product  is to set $\langle{\Omega^\dagger {\bf a}},{{\bf b}}\rangle = \langle{{\bf a}},{\Omega{\bf b}}\rangle $, but this  leads to
$$
\langle{{\bf b}},{{\bf a}}\rangle =\langle{\Omega  {\bf a}}, {\Omega {\bf b}}\rangle \stackrel{?}{=} \langle{\Omega^\dagger \Omega {\bf a}},  {{\bf b}}\rangle 
$$
and  a contradiction: the leftmost expression is antilinear in ${\bf b}$ while the rightmost is linear in ${\bf b}$.   A similar issue leads to 
$$
(\langle {\bf a}| \Omega)|{\bf b}\rangle \ne \langle {\bf a}|(\Omega |{\bf b}\rangle)
$$
and so makes ``matrix elements'' $\langle{{\bf a}}|{\Omega}|{{\bf b}}\rangle $ ambiguous, and    prevents us from defining  the usual  left action of  $\Omega$ on bra vectors $\langle{\bf a}|$.
Instead we have 
$$
\langle{\bf b},{\Omega{\bf a}}\rangle = \langle {\bf a},{\Omega^{-1} {\bf b}}\rangle
= \langle\Omega^{-1} {\bf b}, {\bf a}\rangle ^*.
$$
