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(Note: I have edited the answer to cover all cases.)

Let's call this "property 1". I claim that $\mathcal{O}$ has property 1 if and only if $h(\mathcal{O}_K)$ is odd.

First let me prove it when $\mathcal{O} = \mathcal{O}_K$. Then for a fractional ideal to have norm 1, it must be of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal (here I am tacitly using that ideals in $\mathcal{O}_K$ have unique factorization into prime ideals). Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.

Now consider the general case where $\mathcal{O}$ is an order of index $f$ in $\mathcal{O}_K$. I'll show that $\mathcal{O}$ has property 1 iff $\mathcal{O}_K$ has property 1. If $I$ is a proper fractional $\mathcal{O}$-ideal of norm 1, then $\mathcal{O}_KI$ (the product lattice) has norm 1, since norm is multiplicative. Since the map $\pi_f \colon \mathrm{Pic}(\mathcal{O}) \to \mathrm{Pic}(\mathcal{O}_K)$ sending the class of $I$ to the class of $\mathcal{O}_KI$ is surjective, this shows that if $\mathcal{O}$ has property 1, then so does $\mathcal{O}_K$.

Conversely, suppose $\mathcal{O}_K$ has property 1. Then for any ideal class $[I]$ in $\mathrm{Pic}(\mathcal{O}_K)$, I may choose $I$ to have norm 1 and also to be prime to $f$. Now, one can check that the ideals classes in the pre-image $\pi_f^{-1}([I])$ all have representatives $J$ which are simply index $f$ sublattices in $I$. [Let me not prove this here, but this is exactly what Siegel is doing explicitly in the paper you link to in the comments below. Note, in general not all index $f$ sublattices are proper $\mathcal{O}$-ideals. In fact, the number of such is precisely the size of the kernel of the map $\pi_f$, at least assuming there are no extra units in $\mathcal{O}_K^\times$.] Note that each these $\mathcal{O}$-ideals $J$ has norm 1 as well. So property 1 holds for $\mathcal{O}$.

As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. (And the class number of any sub-order is a multiple of $h(\mathcal{O}_K)$ so it fails for the orders as well). But if $K$ has prime discriminant then indeed the class number is odd.

(Note: I have edited the answer to cover all cases.)

Let's call this "property 1". I claim that $\mathcal{O}$ has property 1 if and only if $h(\mathcal{O}_K)$ is odd.

First let me prove it when $\mathcal{O} = \mathcal{O}_K$. Then for a fractional ideal to have norm 1, it must be of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal (here I am tacitly using that ideals in $\mathcal{O}_K$ have unique factorization into prime ideals). Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.

Now consider the general case where $\mathcal{O}$ is an order of index $f$ in $\mathcal{O}_K$. I'll show that $\mathcal{O}$ has property 1 iff $\mathcal{O}_K$ has property 1. If $I$ is a proper fractional $\mathcal{O}$-ideal of norm 1, then $\mathcal{O}_KI$ (the product lattice) has norm 1, since norm is multiplicative. Since the map $\pi_f \colon \mathrm{Pic}(\mathcal{O}) \to \mathrm{Pic}(\mathcal{O}_K)$ sending the class of $I$ to the class of $\mathcal{O}_KI$ is surjective, this shows that if $\mathcal{O}$ has property 1, then so does $\mathcal{O}_K$.

Conversely, suppose $\mathcal{O}_K$ has property 1. Then for any ideal class $[I]$ in $\mathrm{Pic}(\mathcal{O}_K)$, I may choose $I$ to have norm 1 and also to be prime to $f$. Now, one can check that the ideals classes in the pre-image $\pi_f^{-1}([I])$ all have representatives $J$ which are simply index $f$ sublattices in $I$. [Let me not prove this here, but this is exactly what Siegel is doing explicitly in the paper you link to in the comments below. Note, in general not all index $f$ sublattices are proper $\mathcal{O}$-ideals. In fact, the number of such is precisely the size of the kernel of the map $\pi_f$, at least assuming there are no extra units in $\mathcal{O}_K^\times$.] Note that each these $\mathcal{O}$-ideals $J$ has norm 1 as well. So property 1 holds for $\mathcal{O}$.

As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. But if $K$ has prime discriminant then indeed the class number is odd.

I believe this happens if and only if the class number is odd. For an ideal class to have this property it must be represented by a fractional ideal of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal. Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.

As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. (And the class number of any sub-order is a multiple of $h(\mathcal{O}_K)$ so it fails for the orders as well). But if $K$ has prime discriminant then indeed the class number is odd.

(Note: I have edited the answer to cover all cases.)

Let's call this "property 1". I claim that $\mathcal{O}$ has property 1 if and only if $h(\mathcal{O}_K)$ is odd.

First let me prove it when $\mathcal{O} = \mathcal{O}_K$. Then for a fractional ideal to have norm 1, it must be of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal (here I am tacitly using that ideals in $\mathcal{O}_K$ have unique factorization into prime ideals). Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.

Now consider the general case where $\mathcal{O}$ is an order of index $f$ in $\mathcal{O}_K$. I'll show that $\mathcal{O}$ has property 1 iff $\mathcal{O}_K$ has property 1. If $I$ is a proper fractional $\mathcal{O}$-ideal of norm 1, then $\mathcal{O}_KI$ (the product lattice) has norm 1, since norm is multiplicative. Since the map $\pi_f \colon \mathrm{Pic}(\mathcal{O}) \to \mathrm{Pic}(\mathcal{O}_K)$ sending the class of $I$ to the class of $\mathcal{O}_KI$ is surjective, this shows that if $\mathcal{O}$ has property 1, then so does $\mathcal{O}_K$.

Conversely, suppose $\mathcal{O}_K$ has property 1. Then for any ideal class $[I]$ in $\mathrm{Pic}(\mathcal{O}_K)$, I may choose $I$ to have norm 1 and also to be prime to $f$. Now, one can check that the ideals classes in the pre-image $\pi_f^{-1}([I])$ all have representatives $J$ which are simply index $f$ sublattices in $I$. [Let me not prove this here, but this is exactly what Siegel is doing explicitly in the paper you link to in the comments below. Note, in general not all index $f$ sublattices are proper $\mathcal{O}$-ideals. In fact, the number of such is precisely the size of the kernel of the map $\pi_f$, at least assuming there are no extra units in $\mathcal{O}_K^\times$.] Note that each these $\mathcal{O}$-ideals $J$ has norm 1 as well. So property 1 holds for $\mathcal{O}$.

As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. (And the class number of any sub-order is a multiple of $h(\mathcal{O}_K)$ so it fails for the orders as well). But if $K$ has prime discriminant then indeed the class number is odd.

I believe this happens if and only if the class number is odd. For an ideal class to have this property it must be represented by an element of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal. Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.

As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. (And the class number of any sub-order is a multiple of $h(\mathcal{O}_K)$ so it fails for the orders as well). But if $K$ has prime discriminant then indeed the class number is odd.

I believe this happens if and only if the class number is odd. For an ideal class to have this property it must be represented by a fractional ideal of the form $I/\bar I$, where $\bar I$ is the Galois-conjugate ideal. Since conjugation is inversion in the class group, this means that the ideal class is a square. If all ideal classes are squares then the squaring map is surjective, hence injective, and so the 2-torsion subgroup is trivial. So the group has odd order.

As for your question of how often this happens, the 2-part of the class group is related to the number of primes dividing the discriminant. If, for example, there are at least 2 odd primes $p, q$ dividing the discriminant of $K$, then the 2-part is non-trivial. Indeed, the unique ideal above $p$ is 2-torsion in the class group of the maximal order, but not principal. (And the class number of any sub-order is a multiple of $h(\mathcal{O}_K)$ so it fails for the orders as well). But if $K$ has prime discriminant then indeed the class number is odd.