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# Python 3.8+, 49 byte

```python
lambda c:5**((i:="IVXLCDM".find(c))&1)*10**(i>>1)
```

Testing:

```python
f=lambda c:5**((i:="IVXLCDM".find(c))&1)*10**(i>>1)
print(*map(f,"IVXLCDM"))
```

# 52 bytes

My first attempt.  Really simple and straight forward.  If the rules would allow `1000.0` as result for `M` then it could be written as `1e3`, making this solution one byte smaller.

```python
lambda n:[1,5,10,50,100,500,1000]["IVXLCDM".find(n)]
```