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Here is a simple answer to the first part of the question. See Peter's comment below for an explanation of why the same example also answers the second part of the question.

One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories.

For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},\leq\rangle\qquad \langle\mathbb{Q}^{\geq 0},\leq\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not.

But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x\leq y$ there is a unique morphism from $x$ to $y$, which is the identity morphism when $x=y$.

The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not.

One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories.

For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},\leq\rangle\qquad \langle\mathbb{Q}^{\geq 0},\leq\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not.

But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x\leq y$ there is a unique morphism from $x$ to $y$, which is the identity morphism when $x=y$.

The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not.

This example answers directly the first part of the question. But actually it answers also the second part of the question, as explained in Peter's comment.

Here is a simple answer to the first part of the question. One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories.

For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},\leq\rangle\qquad \langle\mathbb{Q}^{\geq 0},\leq\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not.

But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x\leq y$ there is a unique morphism from $x$ to $y$, which is the identity morphism when $x=y$.

The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not.

Here is a simple answer to the first part of the question. See Peter's comment below for an explanation of why the same example also answers the second part of the question.

One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories.

For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},\leq\rangle\qquad \langle\mathbb{Q}^{\geq 0},\leq\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not.

But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x\leq y$ there is a unique morphism from $x$ to $y$, which is the identity morphism when $x=y$.

The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not.

Here is a simple answer to the first part of the question. One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories.

For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},<\rangle\qquad \langle\mathbb{Q}^{\geq 0},<\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not.

But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x<y$ there is a unique morphism from $x$ to $y$, plus the identity morphisms.

The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not.

Here is a simple answer to the first part of the question. One can take some of the standard violations of CSB with other kinds of mathematical structures and transfer them to categories.

For example, with linear orders, we have the two linear orders $$\langle\mathbb{Q},\leq\rangle\qquad \langle\mathbb{Q}^{\geq 0},\leq\rangle,$$ which each order-embed into each other, but they are not isomorphic, since the latter has a minimal element and the former does not.

But any linear order can be viewed as a category, where one takes the nodes of the order as objects, and whenever $x\leq y$ there is a unique morphism from $x$ to $y$, which is the identity morphism when $x=y$.

The order embeddings give rise to embedding functors in each direction for the categories, but these two categories are not isomorphic, since the latter one has an initial object, but the former does not.