Can light travel slower than the maximum?
The velocity of light is locally invarient, that is if you measure it at your position you will always get the value $c$. The other answers explain why in your example of the astronaut with the torch we would still measure the speed of light to be $c$.
However in General Relativity the speed of light is not globally invarient. For example if your astronaut is hovering just above the event horizon of a black hole and shines his torch outwards we would see the light initially travelling at less than $c$ and accelerating towards $c$ as it gets farther from the black hole. At the event horizon itself we would see the speed of the light slow to zero. See this Wikipedia article for some useful links, though the article itself doesn't say much more than I've said here.
You need to be a bit cautious attaching physical relevance to this, because any velocities we measure is not invarient quantities in GR. The proper velocity of light is an invariant and remains $c$ even at the black hole event horizon. Still this is an example of how the speed of light can be changed and the light can be affected by gravity.
Not only is the constant nature of the speed of light guaranteed by theory, it is also shown experimentally. In fact, as you may know, it was the experimental discovery that the speed of light is constant irrespective of the (inertial) frame of reference which formed the inspiration for the development of special relativity by Albert Einstein.
Mathematically, purely from the relativistic formula for velocity-addition it can be seen that the light would still travel at a speed $c$ in a vacuum. Indeed, say the astronaut has a speed of $-v$ with respect to the ground. He observes the light leaving his flashlight with a speed of $u = +c$ (with respect to him).$^1$ The relativistic theory then tells us the light is travelling at a speed $s$ with respect to the ground, given by
$$\begin{align} s &= \frac{(-v)+(+c)}{1+\frac{(-v)(+c)}{c^2}} \\ \\ &= \frac{c-v}{1-\frac{v}{c}} \\ \\ &= \frac{c-v}{\frac{c-v}{c}} \\ \\ &= \frac{c-v}{c-v}c \\ \\ s &= c. \end{align}$$
You can walk at whatever speed you like, you'll never get a different result. Except if you insert $-c$ instead of $-v$, then the answer is undefined - however, you do still get $c$ if you calculate it using limits.
$^1$ Why can we say this? Well, remember that in the reference frame of the astronaut, he himself is not moving at all and he would expect the light to leave his flashlight at a speed of $c$ (w.r.t. him). This becomes clearer when you replace the flashlight with a small cannon and the light with a little ball.
Suppose you can set the exit speed for this ball out of the cannon. Say you set it at $v$. When you walk around carrying this small cannon, you expect the ball to still leave the cannon at a speed $v$ with respect to you when you fire it. Classically, if you’re walking at a speed $u$ with respect to the ground, the ball will leave the cannon at a speed $u+v$ with respect to the ground.
Relativistically, the only thing that is invalid in all the above is this simple summation of speeds. You need the formula I used in the main text of this post, which reduces to the simple summation if both $u$ and $v$ are much smaller than $c$. So it’s perfectly alright to say that the light leaves the astronaut’s flashlight at a speed $c$ w.r.t. the astronaut, even without knowing about the constancy of the speed of light in a vacuum.
What if an astronaut during a spacewalk switches on a flashlight towards the opposite direction he is moving. Will that light travel slower than normal?
In Newtonian mechanics, object A is moving at velocity $u$ relative to B, and B is moving at $v$ relative to C, then A's velocity relative to C is simply $u+v$. But in special relativity, the equation is $(u+v)/(1+uv)$ (assuming that $u$ and $v$ are both relative to the speed of light, i.e., the units are such that $c=1$). If $u=1$, then the result of the combination of velocities is still 1.