The answer is yes.

Consider the collection $\mathcal C$ of hypergraphs of the following form. They have underlying set $\omega$ as the vertices, the natural numbers. The finite edges in the hypergraph are all and only the sets of the form $\{0,1,\ldots,n\}$. And then the hypergraph can have any desired collection of infinite edge sets.

Since there are $2^{\aleph_0}$ many infinite subsets of $\omega$, it follows that there are $2^{2^{\aleph_0}}$ many members of $\mathcal C$. And I claim that they are all pairwise non-isomorphic. To see this, note first that $\{0\}$ is the only edge with one member, and so the isomorphism must fix $0$. Similarly, $\{0,1\}$ is the only edge with two members, and so the isomorphism must fix $1$. And so on. The finite edges force the isomorphism to fix every individual vertex, and so non-identical members of $\mathcal C$ will be non-isomorphic.

EDIT   Let $\ F:=\ \{\{1\ \ldots\ n\}: n\in\mathbb N\},\ $ and $\ D:=\{X\in 2^\mathbb N : |X|=|\mathbb N|\}. $ Then define: $$ C\ :=\ \{ F\cup G:\ G\subseteq D\} $$

The answer is yes.

Consider the collection $\mathcal C$ of hypergraphs of the following form. They have underlying set $\omega$ as the vertices, the natural numbers. The finite edges in the hypergraph are all and only the sets of the form $\{0,1,\ldots,n\}$. And then the hypergraph can have any desired collection of infinite edge sets.

Since there are $2^{\aleph_0}$ many infinite subsets of $\omega$, it follows that there are $2^{2^{\aleph_0}}$ many members of $\mathcal C$. And I claim that they are all pairwise non-isomorphic. To see this, note first that $\{0\}$ is the only edge with one member, and so the isomorphism must fix $0$. Similarly, $\{0,1\}$ is the only edge with two members, and so the isomorphism must fix $1$. And so on. The finite edges force the isomorphism to fix every individual vertex, and so non-identical members of $\mathcal C$ will be non-isomorphic.

EDIT   Let $\ F:=\ \{\{1\ \ldots\ n\}: n\in\mathbb N\},\ $ and $\ D:=\{X\in 2^\mathbb N : |X|=|\mathbb N|\}. $ Then define: $$ C\ :=\ \{ F\cup G:\ G\subseteq D\} $$ The Dominic's isomorphisms are reduced to just one identity ismorphism per selected hypergraph, i.e. there are no non-trivial Domininc's isomorphisms between members of $\ C$.

The answer is yes.

Consider the collection $\mathcal C$ of hypergraphs of the following form. They have underlying set $\omega$ as the vertices, the natural numbers. The finite edges in the hypergraph are all and only the sets of the form $\{0,1,\ldots,n\}$. And then the hypergraph can have any desired collection of infinite edge sets.

Since there are $2^{\aleph_0}$ many infinite subsets of $\omega$, it follows that there are $2^{2^{\aleph_0}}$ many members of $\mathcal C$. And I claim that they are all pairwise non-isomorphic. To see this, note first that $\{0\}$ is the only edge with one member, and so the isomorphism must fix $0$. Similarly, $\{0,1\}$ is the only edge with two members, and so the isomorphism must fix $1$. And so on. The finite edges force the isomorphism to fix every individual vertex, and so non-identical members of $\mathcal C$ will be non-isomorphic.

The answer is yes.

Consider the collection $\mathcal C$ of hypergraphs of the following form. They have underlying set $\omega$ as the vertices, the natural numbers. The finite edges in the hypergraph are all and only the sets of the form $\{0,1,\ldots,n\}$. And then the hypergraph can have any desired collection of infinite edge sets.

Since there are $2^{\aleph_0}$ many infinite subsets of $\omega$, it follows that there are $2^{2^{\aleph_0}}$ many members of $\mathcal C$. And I claim that they are all pairwise non-isomorphic. To see this, note first that $\{0\}$ is the only edge with one member, and so the isomorphism must fix $0$. Similarly, $\{0,1\}$ is the only edge with two members, and so the isomorphism must fix $1$. And so on. The finite edges force the isomorphism to fix every individual vertex, and so non-identical members of $\mathcal C$ will be non-isomorphic.

EDIT   Let $\ F:=\ \{\{1\ \ldots\ n\}: n\in\mathbb N\},\ $ and $\ D:=\{X\in 2^\mathbb N : |X|=|\mathbb N|\}. $ Then define: $$ C\ :=\ \{ F\cup G:\ G\subseteq D\} $$