Q:

Python: What does *= mean?

In Python, what does it mean when *= is used.  For example:
for i in xrange(len(files)):
    itimes[i,:,:] *= thishdr["truitime"]

A:

As others have explained, this is roughly an equivalent to:
[object] = [object] * [another_object]

However, it's not exactly the same. Technically, the above calls the __mul__ function, which returns a value, and reassign it back to the name.
For example, we have an object A and multiplying it with B. The process is something like this:
> Call the __mul__ function of object A, 
> Retrieve the new object returned.
> Reassign it to the name A.

Looks simple. Now, by doing *= we're not invoking the method __mul__, but instead __imul__, which will attempt to modify itself. The process is something like this:
> Call the __imul__ function of object A,
> __imul__ will change the value of the object, it returns the modified object itself
> The value is reassigned back to the name A, but still points to the same place in memory.

With this, you're modifying it in-place, not creating a new object. 
So what? It looks the same..
Not exactly. If you replaces an object, you created a new place for it in memory. If you modify it in-place, the object location in the memory will always be the same. 
Take a look at this console session:
>>> a = [1, 2, 3]
>>> b = a
>>> c = 10
>>> a = a * c
>>> print a
[1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3]
>>> b
[1, 2, 3]

If we check the memory address:
>>> id(a) == id(b)

False
Using this, the value of b is unchanged, since a is now just pointing to a different place. But using *=:
>>> a = [1, 2, 3]
>>> b = a
>>> c = 10
>>> a *= c
>>> b
[1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3]

And if we check the memory address:
>>> id(a) == id(b)
True

The operation affects b as well. This can be tricky and leading to confusing behaviours, sometimes. But once you understand it, it would be easy to handle.
Hope this helps!

A:

It just means "[expression on the left] = [itself] * [expression on the right]":
itimes[i,:,:] *= thishdr["truitime"]
is equivalent to
itimes[i,:,:] = itimes[i,:,:] * thishdr["truitime"]

