What is the n factor of h2o2 undergoing disproportionation reaction liberating oxygen
We can split the reaction
$$\ce{H2O2 -&gt; H2O + O2}$$
into the respective reduction and oxidation half-reactions.
$$\ce{H2O2 -&gt; O2 + 2H+ + 2e-}$$
$$\ce{H2O2 + 2H+ + 2e- -&gt; 2H2O}$$
Since the n-factor of $\ce{H2O2}$ for both these half-reactions is 2, the n-factor is:
$$\frac{1}{n_f} = \frac{1}{2} + \frac{1}{2} = 1$$
$$n_f = 1$$
Based on the comments under your post I believe you have gotten this far, but some sources claim that the n-factor for this reaction is 2. Well, how do you verify what is correct? In such a case, I'd simply write the balanced chemical reaction and verify whether the law of chemical equivalence holds or not.
$$\ce{2H2O2 -&gt; 2H2O + O2}$$
The n-factor for $\ce{O2}$ should be $2$ as can be seen in the first half-reaction, and for $\ce{H2O}$ the n-factor should be $1$, since we have $2$ moles of $\ce{e-}$ in the second half-reaction and $2$ moles of $\ce{H2O}$, and n-factor is defined per-mole of the substance.
From the law of chemical equivalence, the number of equivalents of $\ce{H2O2}$, $\ce{H2O}$ and $\ce{O2}$ should all be equal. If the claim I make above that the n-factor of $\ce{H2O2}$ is $1$ is true, this should hold.
Plugging in the moles from the balanced chemical equation and using the n-factors described above,
Equivalents of $\ce{H2O2} = 2 \times 1$, $\ce{H2O} = 2 \times 1$, $\ce{O2} = 1 \times 2$ (moles $\times$ n-factor)
Thus the n-factor is indeed $1$. The n-factor of $\ce{H2O2}$ for the half-reactions is $2$.
(Note that all the n-factors above are with respect to the reaction $\ce{H2O2 -&gt; H2O + O2}$ and they (most probably) will change for a different reaction)
Update (the above part answered the question before the edit):
From [1]:

The &quot;volume strength&quot; of hydrogen peroxide is defined as the number of times its own volume of oxygen a sample of hydrogen peroxide solution will evolve if decomposed naturally.
$$ \ce{H2O2 -&gt; H2O + O2}$$

As you've mentioned in your question,
$$\pu{Normality = \frac{Volume~strength}{5.6}}$$
and
$$\pu{Molarity = \frac{Volume~strength}{11.2}}$$
This does not match with the reasoning that the n-factor of $\ce{H2O2}$ for the above reaction is $1$, which would imply that the normality and molarity should be equal.
From [2]:

Now, equivalent weight of $\ce{H2O2} = \frac{34}{2} = 17$


A (N) $\ce{H2O2}$ solution contains 17 gms of $\ce{H2O2}$ per litre.

So in fact when we are relating the normality of an $\ce{H2O2}$ solution to it's volume strength, we are not referring to the normality of $\ce{H2O2}$ in the decomposition reaction. You would be right in saying that the normality of $\ce{H2O2}$ in the decomposition reaction is equal to its molarity.
One way of justifying the n-factor of $2$ of an $\ce{H2O2}$ solution (not undergoing any reaction) is due to the $-2$ charge on the $\ce{O2^{2-}}$ peroxide ion.
I can't really explain why this has been done, since it would make just as much sense to consider the n-factor of $\ce{H2O2}$ in the decomposition reaction, it is just a historical fact. Since n-factor is not a constant for a particular compound, it should be really important to specify context about normality when we relate it to volume strength, which is rarely done when the relation is taught which is what leads to this confusion. Both the n-factor's are correct in different contexts, so I would just recommend sticking to the n-factor of $2$ when talking about volume strength since that is what you have been taught.
[1]: Chemical Calculations by King A archive.org
[2]: An Introduction To Chemistry Vol. 1 Ed. 3rd by Das, Ranajit archive.org