I agree with Wolphram jonny's answer that even if your fist has the same momentum in both cases, the force your fist exerts on the feather will be less than the force on the glass, and in each case the force your fist exerts is matched by an equal and opposite force from the object you're punching. But I thought I'd also add a simple argument for why this should be the case. Suppose we simplify things by assuming the collision between your fist and the object is an elastic one--no kinetic energy from your fist is dispersed as heat or sound waves. In this case, by combining an equation expressing conservation of momentum with an expression expressing conservation of linear kinetic energy, it's possible to derive these equations for the case of a head-on collision where object 2 is initially at rest:
$v^{\prime}_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1$
$v^{\prime}_2 = \frac{2m_1}{m_1 + m_2} v_1$
Here $m_1$ is the mass of the initially moving object, in this case your fist, while $m_2$ is the mass of the object initially at rest, the feather or glass. $v_1$ is the initial velocity of your fist before the collision, $v^{\prime}_1$ is the velocity of your fist after the collision, and $v^{\prime}_2$ is the velocity of the feather or glass after the collision. So the change in momentum of your fist would be $\Delta p_1 = m_1 (v^{\prime}_1 - v_1)$, while the change in momentum of the glass or feather would be $\Delta p_2 = m_2 v^{\prime}_2$ (A little algebra will verify that $\Delta p_2 = - \Delta p_1$ as required by conservation of momentum).
Since the change in momentum is the impulse, for the period $\Delta t$ the fist and the object are actually in contact, the average force on the object $F_{avg}$ will obey $F_{avg} \Delta t = \Delta p_2$. So, this gives us:
$F_{avg} \Delta t = \frac{2m_1 m_2}{m_1 + m_2} v_1$
During the period of an elastic collision, the material of the object being punched briefly gets compressed like a spring, then expands like a compressed spring with one end free, and that (along with your fist doing the same thing) is what causes it to fly off (see this pdf for some detailed analysis of collisions with equations). So the period $\Delta t$ between when your fist starts to contact the object and when the object springs off depends on the material the object is made of, but as a further simplification let's assume we are punching two objects made out of the same type of material, so $\Delta t$ is the same in both cases, but with the two objects having different values for the mass $m_2$. In that case, you can see from the above equation that the value of $F_{avg}$ will depend a great deal on the value of $m_2$. In the limit as $m_2$ goes to zero, $F_{avg} \Delta t$ goes to zero as well. And in the limit as $m_2$ becomes very large compared to $m_1$, $\frac{2m_1 m_2}{m_1 + m_2} v_1$ becomes very close to $\frac{2m_1 m_2}{m_2} v_1 = 2 m_1 v_1$, twice the original momentum of the fist approaching it. So, you can see that if you punch different objects with your fist having the same momentum $p_1$, $F_{avg} \Delta t$ can have any value between $0$ and $2p_1$, depending on the mass of the object you're punching.
In reality a more elastic object like feather should remain in contact with your fist for a greater time $\Delta t^{\prime}$ than the time $\Delta t$ for a more rigid object like a sheet of glass. So given that the equation indicates the average force multiplied by the time interval is greater for the glass than the feather, if the average collision force is $F^{\prime}_{avg}$ for the feather and $F_{avg}$ for the glass, we have $F_{avg} \Delta t > F^{\prime}_{avg} \Delta t^{\prime}$ and $\Delta t < \Delta t^{\prime}$, which implies that $F_{avg}$ for the glass must be greater than $F^{\prime}_{avg}$ for the feather by an even larger amount than if the time intervals were equal.
Edited to add: Bert and bobie's answers appear to deny that the third law is applicable to cases where one object breaks through another one, with Bert saying "you are right, the reaction of a (window) glass is not always equal to the action", and bobie saying " If we punch it with 200N, it cannot give an equal 'reaction'. If we call 'action' the punch, the 'reaction' of the glass cannot exceed 100N." I suppose it's possible they might just be saying that if you punched a glass with enough momentum so your fist would exert 200N if it was hitting a less breakable force meter instead of a piece of glass, then the glass will exert less than 200N on your fist, without them actually claiming that your fist was exerting 200N on the glass in this case. But I have asked both of them for clarification on this point in comments (my comments on bobie's post were apparently deleted) and neither of them have been willing to discuss it. This makes me suspect they probably are just disagreeing with the point I (and others) have made that the forces between two bodies are always equal and opposite, though the value of these equal and opposite forces can vary depending on what it is you're punching.
If they are denying that the forces are equal and opposite even for examples like an object smashing through a glass, I just wanted to add that this is definitely not the mainstream physics view. In my comments to Bert I pointed to the snippets from the solutions manual to a college physics textbook here and here (and some additional snippets which can be seen googling specific phrases) which pose the following question and answer:
A brick hits a window and breaks the glass. Since the brick breaks the
glass,
(a) the force on the brick is greater than the force on the glass,
(b) the force on the brick is equal to the force on the glass,
(c) the force on the brick is smaller than the force on the glass,
(d) the force on the brick is in the same direction as the force on
the glass.
Solution: According to Newton's third law, the answer is (b). Why
isn't answer (c) correct since the brick breaks the glass? It takes a
considerably smaller force to break a glass than to break a brick.
When the brick hits the glass, they exert equal but opposite forces on
each other, say 150 N. It may take only 100 N to break the glass and
1000 N to break the brick. So the glass breaks and the brick remains
basically undamaged
Likewise, from the same page, there's this snippet:
There is another misconception concerning the third law. The third law
states that the two forces are equal no matter what. For example, an
egg and a stone collide with each other, the egg breaks and the stone
is intact. Since the egg breaks, we often conclude that the force by
the stone on the egg is greater than the force by the egg on the
stone. This is not so. The forces are always equal. The egg breaks
because it is simply easier to break.
More generally, if you go to google books and perform this search for "Newton's third law" along with the word "always", you can find a large number of textbooks saying that the forces between two interacting objects are always equal and opposite, with no exceptions noted for breakage or any other classical scenario. And note that when Bert and bobie suggest the forces would be unbalanced, they never cite any physics expert making this claim, nor do they offer any mathematical derivation of this claim or any experimental evidence to believe it's true.
