Transformation of states and operators under Quantum Lorentz Transformation For any Homogenous Proper Orthochronous Lorentz Transformation $$x\to\Lambda x$$ , there is a unitary linear operator $$U(\Lambda)$$ [Bogoliubov(1980),Srednicki(2007),Weinberg(1995)] which can be applied to either the state vector $$\left|\psi\right> \to U(\Lambda)\left|\psi\right>$$ of the field (S-Picture) or the linear operators corresponding to the observables(H-Picture) $$L \to U^{-1}(\Lambda)LU(\Lambda)$$ [Bogoliubov 1980]. Now,if we consider the transformation $$ x \to x'=\Lambda x \to x''=\tilde \Lambda x'$$ in the S-picture the transformation is following: $$\left|\psi\right> \to U(\Lambda)\left|\psi\right> \to U(\tilde \Lambda)U(\Lambda)\left|\psi\right>$$ and the operators corresponding to some physical quantity is fixed (assuming no interaction) whereas the H-Picture the transformation is the following:
$$ L \to U^{-1}(\Lambda)LU(\Lambda) \to U^{-1}(\tilde \Lambda)U^{-1}(\Lambda)LU(\Lambda)U(\tilde \Lambda)$$ and the state vector remains fixed for all time (assuming no interaction).
For S-picture the expectation value is: $$ \left<\psi\right|U^{-1}(\Lambda)U^{-1}(\tilde\Lambda)LU(\tilde\Lambda)U(\Lambda)\left|\psi\right>$$
And in the H-picture the value is : $$\left<\psi\right|U^{-1}(\tilde\Lambda)U^{-1}(\Lambda)LU(\Lambda)U(\tilde\Lambda)\left|\psi\right>$$
Now for the equivalence of both picture, we need the same expectation value of the observable in both picture which can be done by assuming the following: $$U(\Lambda)U(\tilde\Lambda) = U(\tilde\Lambda)U(\Lambda) \tag{1}\label{1}$$
My question is whether equation \eqref{1} is true  and  if not, what additional condition ,besides linearity and unitarity, enable $$U(\Lambda)$$ to give $$\left<\psi\right|U^{-1}(\tilde\Lambda)U^{-1}(\Lambda)LU(\Lambda)U(\tilde\Lambda)\left|\psi\right>=\left<\psi\right|U^{-1}(\Lambda)U^{-1}(\tilde\Lambda)LU(\tilde\Lambda)U(\Lambda)\left|\psi\right> \tag{2}$$ for any observable L
 A: I guess your transformation behaviour of $L$ is wrong. Given a transformation $U$ we have that it acts on a state $|\psi>$ as $U|\psi>$. From that one can deduce the transformation behaviour of an operator $A$: We have that $A|\psi>$ transforms as $UA|\psi>$ which is the same as $UAU^{-1}U|\psi>$ such that the action of the operator on the new transformed state $U|\psi>$ is given $UAU^{-1}$, which is defined as the transform operator by looking at the graph of the operator. Hence you obtain when deriving the expression of the expectation value you do not arrive at different orderings regarding the transformation parameter. You obtain the same ordering.
So this is the problem and not regarding to the difference of the Heisenberg and Schrödering picture.
The only difference between the Schrödering and Heisenberg picture is that all operators in the Heisenberg Picture have been evolved with the INVERSE time operator $U^{-1}$ (I guess this is the source of your confusion with the transformation of L) such that according to the above discussion we have that on a time-dependent state $|\psi(t)> \to U^{-1}(t)|\psi(t)>=|\psi(0)>$ so the new state is time-independent, while the operator $A$ now is given by $U^{-1}(t)AU(t)$.
UPDATE:
Since you just need to transform time to go from the Heisenberg picture to the Schrödering picture, you basically look at the subgroup of spacetime-translations, which is abelian. Hence, equation (1) holds for that subgroup (ofc not for the hole lorentz group).
A: Equation $(1)$ is false, because the Lorenz group is not abelian and there its representations are not abelian either (at least the one useful in QFT). There is no reason for $(2)$ to hold either.
This shows that what you wrote for the Heisenberg picture is not actually a representation of the Lorenz group. Instead, it should be :
$$L\to U(\Lambda)LU(\Lambda)^{-1}$$
which behaves nicely under two successive Lorenz transformations :
$$L \to U(\Lambda)LU^{-1}(\Lambda)\to U(\Lambda')U(\Lambda) LU(\Lambda)^{-1}U(\Lambda')^{-1} = U(\Lambda'\Lambda)LU(\Lambda'\Lambda)^{-1}$$
A: Shortening the notation the LHS of your formula (2) can be written as
$$
\langle\psi,\tilde U^{-1}U^{-1}\,L\,U\,\tilde U\,\psi\rangle\,.
$$
Because $U$ and $\tilde U$ are unitary this equals
$$
\langle U\,\tilde U\psi,L\,U\,\tilde U\,\psi\rangle=\langle\psi,L\psi\rangle\,.
$$
The last equals sign is true because $U\tilde U$ is unitary. Obviously we don't need assumption (1).
