Cloned estimators have identical randomness but different RNG instances

### Describe the bug

Cloned estimators have identical randomness but different RNG instances. According to documentation, it should be the other way around: different randomness but identical RNG instances.

Related #25395 

The User Guide [says](https://scikit-learn.org/stable/common_pitfalls.html#controlling-randomness):
> For an optimal robustness of cross-validation (CV) results, pass RandomState instances when creating estimators

> ```python
> rf_inst = RandomForestClassifier(random_state=np.random.RandomState(0))
> cross_val_score(rf_inst, X, y)
> ```
> ...
> Since `rf_inst` was passed a `RandomState` instance, each call to `fit` starts from a different RNG. As a result, the random subset of features will be different for each folds

In regards to cloning, the same reference says:
> ```python
> rng = np.random.RandomState(0)
> a = RandomForestClassifier(random_state=rng)
> b = clone(a)
> ```
> Moreover, `a` and `b` will influence each-other since they share the same internal RNG: calling `a.fit` will consume `b`’s RNG, and calling `b.fit` will consume `a`’s RNG, since they are the same. 

The actual behaviour does not follow this description. 

### Steps/Code to Reproduce

```python
import numpy as np
from sklearn import clone
from sklearn.datasets import make_classification
from sklearn.model_selection import cross_validate
from sklearn.ensemble import RandomForestClassifier

rng = np.random.RandomState(0)
X, y = make_classification(random_state=rng)
rf = RandomForestClassifier(random_state=rng)

d = cross_validate(rf, X, y, return_estimator=True, cv=2)
rngs = [e.random_state for e in d['estimator']]
# estimators corresponding to different CV runs have different but identical RNGs:
print(rngs[0] is rngs[1]) # False
print(all(rngs[0].randint(10, size=10) == rngs[1].randint(10, size=10))) # True

rf_clone = clone(rf)
rngs = [rf.random_state, rf_clone.random_state]
print(rngs[0] is rngs[1]) # False
print(all(rngs[0].randint(10, size=10) == rngs[1].randint(10, size=10))) # True
```

### Expected Results

True False True False

### Actual Results

False True False True

### Versions

```shell
Tested on a two-week-old dev build and also on the following version (Kaggle)

System:
    python: 3.7.12 | packaged by conda-forge | (default, Oct 26 2021, 06:08:53)  [GCC 9.4.0]
executable: /opt/conda/bin/python3.7
   machine: Linux-5.15.90+-x86_64-with-debian-bullseye-sid

Python dependencies:
          pip: 22.3.1
   setuptools: 59.8.0
      sklearn: 1.0.2
        numpy: 1.21.6
        scipy: 1.7.3
       Cython: 0.29.34
       pandas: 1.3.5
   matplotlib: 3.5.3
       joblib: 1.2.0
threadpoolctl: 3.1.0

Built with OpenMP: True
```

COMMENT:
Right, I had more things but its complicated and of course mapping out things is important.  I suppose I have a few questions (opinions):
* I don't really like the idea of `rng=integer` and `rng=np.random.default_rng(integer)` behaving different (they can be result in different values, but not in subtly diffrerent behavior on cloning).
* I don't really like the idea that users might use the same random numbers twice unless they are explicit about it.  With NumPy itself this is almost impossible.
* However, there seems to be a thing around `CV.split()` where doing that may even be the clearly expected thing.
  * Question: Could there be a different way to spell this?  `CV.rewind()`, or `CV(rng=1234, repeat=True)` or...?

The thing about `CV` and estimators being in a different boat seems potentially super tricky, and yes, maybe cloning actually cloning the state too makes that work.  But, maybe it is worth considering if there are orthogenal solutions?

Labels: Bug, Documentation, Needs Decision