How do we perform 'time' translation in Euclidean QFT? If we have an operator in a $1+1$ dimension QFT then we get the Hamiltonian, which comes from and generates translations in the $t$ direction and a momentum operator which comes from and generates translations in the $x$ direction.
If we have an operator, in the Minkowski theory,
$$
O(x,t)=e^{-iHt-iPx}O(0,0)e^{+iHt+iPx}.
$$
If we analytically continue to $t\to -i \tau$ then we get:
$$
O(x,\tau)=e^{-H\tau-iPx}O(0,0)e^{+H\tau+iPx}.
$$
Is it still the case that the Hamiltonian generates time, $\tau$, translations, this operator is no longer unitary and so doesn't look like a symmetry transfomation.
 A: This is an interesting point that I've seen come up on the context of conformal field theories, and the conformal booststrap specifically. In this context at least, the manipulations you're thinking about are related to a property known as reflection positivity.
A lot of information on this and the surrounding area can be found here, and specifically in section 7. But let me try and give a more focused response to your particular question here.
Suppose we have some operator $O_L(t,\boldsymbol x)$ in a Lorentzian theory which we would like to continue to a Euclidean signature. Then as you've pointed out, we must have
$$
O_L(t,\boldsymbol x)=e^{iHt-i\boldsymbol x\cdot\boldsymbol P}O_L(0,0)e^{-iHt+i\boldsymbol{x}\cdot\boldsymbol{P}}.
$$
By Wick rotating to Euclidean time $\tau=it$ we define a new operator in the Euclidean theory by $O_E(\tau,\boldsymbol x)=O_L(-i\tau,\boldsymbol x)$ because remember, we are not really defining a new set of operators, but continuing the ones we already have.
With this, the Euclidean time operator must be expressible in terms of the same(!) Hamiltonian and momentum operators as
$$
O_E(\tau,\boldsymbol x)=O_L(-i\tau,\boldsymbol x)=e^{H\tau-i\boldsymbol{x}\cdot\boldsymbol P}O_L(0,0)e^{-Ht+i\boldsymbol x\cdot\boldsymbol P}=e^{H\tau-i\boldsymbol{x}\cdot\boldsymbol P}O_E(0,0)e^{-Ht+i\boldsymbol x\cdot\boldsymbol P}
$$
where the final equality follows from $\tau=t$ in the special case where both are zero. In this sense, we would say that $e^{-H\tau}$ is the generator of translations in Euclidean time.
You are also right that this operator is no longer unitary, but what we can note is that it's only not unitary because the conjugation hits the $i$ in the time argument at the same time as the operators. The issue at the end of the day comes down to the order of operations: the conjugate of the continued operator is not the continuation of the conjugated operator. So in the Euclidean theory we are free to define two different kinds of Hermitian conjugation, one which is the conjugate of the continued operator which I will denote by $O_E(\tau,\boldsymbol x)^\dagger$ and the continuation of the conjugated operator which I will denote by $O_E^\dagger(\tau,\boldsymbol x)$.
With just a little bit of effort, we can see that, while these two notions of conjugation are not the same, they are related. Specifically, $O_E(\tau,\boldsymbol x)^\dagger=O_E^\dagger(-\tau,\boldsymbol x)$. This is where the "reflection" in "reflection positivity" comes from. Just to avoid leaving things on a cliffhanger, the "positivity" part comes from noting that $\langle\psi|O_E^\dagger(\tau)O_E(-\tau)|\psi\rangle\geq0$ is, using the notations I have introduced, precisely the statement that for any state $|\psi^\prime\rangle=O_L(t)|\psi\rangle$ satisfies $\langle\psi^\prime|\psi^\prime\rangle\geq0$ in a unitary theory.
