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:
I think you are right Nicolas :-/ Sorry.

Having thought about my previous comment regarding JAX Keys and splitting and having a discussion with @seberg I think I am on the cusp of changing my mind again compared to what I wrote :-/

Sebastian pointed out that either option (`fit` modifies the state passed in/`fit` leaves it unmodified) can be a foot-gun. If most people want the two RFs from my example to be independent then it is annoying to make them `split` the key every time (and people will forget to do so, hence foot-gunning themselves). Same argument applies in reverse if most users want the estimators to see the same randomness.

We can make both options happen. One by teaching the minority that they need to call `split` (default is same randomness), the other by teaching the minority that they need to `copy` (default is not the same randomness). I think the important thing is that we don't end up having to teach the majority about an extra step.

Labels: Bug, Documentation, Needs Decision