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For simplicity, I will assume that all manifolds are connected and closed.

An obvious sufficient condition is that the for each pair of embeddings, their images are either disjoint or equal.

To see that this is also necessary, suppose that the two embeddings $f\colon M \hookrightarrow N,\, f'\colon M' \hookrightarrow N$ of copies of $M$ intersect in a submanifold $Q$. Note that $Q = f^{-1}(M') \subseteq M$ can be viewed as a closed submanifold of $M$. This implies that either $Q=M=M'$, or $Q$ has dimension less than that of $M$. In this latter case, the union $f(M)\cup f'(M')$ will not be a submanifold (in each neighbourhood of a point of $Q$ there is a double-point singularity of the immersion $f\sqcup f'$).

The same argument extends to disconnected manifolds and shows that the images of each connected component must be pairwise equal or disjoint.

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For simplicity, I will assume that all manifolds are connected and closed.

An obvious sufficient condition is that the for each pair of embeddings, their images are either disjoint or equal.

To see that this is also necessary, suppose that the two embeddings $f\colon M \hookrightarrow N,\, f'\colon M' \hookrightarrow N$ of copies of $M$ intersect in a submanifold $Q$. Note that $Q = f^{-1}(M') \subseteq M$ can be viewed as a closed submanifold of $M$. This implies that either $Q=M=M'$, or $Q$ has dimension less than that of $M$. In this latter case, the union $f(M)\cup f'(M')$ will not be a submanifold (in each neighbourhood of a point of $Q$ there is a double-point singularity of the immersion $f\sqcup f'$).

The same argument extends to disconnected manifolds and shows that the images of each connected component must be pairwise equal or disjoint.