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One necessary condition is that the center of $G$ needs to act trivially in $V$ for $\mathrm{Hom}(W, W \otimes V)$ to ever be non-trivial. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)

I don't know if this is also sufficient. This boils down to looking at groups with trivial center.

In any case, you originally asked about finite subgroups of SU(2) and its fundamental representation. These are all classified, and none of them have trivial center, so all satisfy this property. Proof of the last bit: any subgroup $G$ of SU(2) gives a subgroup $\overline{G}$ of SO(3) by projection. If $\overline{G}$ has even order, it has an element of order 2, which necessarily lifts to an element of order 4 whose square is $-1$, so $-1 \in G$. The only subgroups of SO(3) of odd order are the odd cyclic groups, which lift to Abelian groups. Some references are is these notes by Dolgachev or an earlier MO question.

One necessary condition is that the center of $G$ needs to act trivially in $V$ for $\mathrm{Hom}(W, W \otimes V)$ to ever be non-trivial. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)

I don't know if this is also sufficient. This boils down to looking at groups with trivial center.

In any case, you originally asked about finite subgroups of SU(2) and its fundamental representation. These are all classified, and none of them have trivial center, so all satisfy this property. Proof of the last bit: any subgroup $G$ of SU(2) gives a subgroup $\overline{G}$ of SO(3) by projection. If $\overline{G}$ has even order, it has an element of order 2, which necessarily lifts to an element of order 4 whose square is $-1$, so $-1 \in G$. The only subgroups of SO(3) of odd order are the odd cyclic groups, which lift to Abelian groups. Some references are is these notes by Dolgachev or an earlier MO question.

One necessary condition (that suffices in your SU(2) case, and all others I could think of) is that the center of $G$ needs to act trivially in $V$ for $\mathrm{Hom}(W, W \otimes V)$ to ever be non-trivial. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)

I don't know if this is also sufficient. This boils down to looking at groups with trivial center.

One necessary condition is that the center of $G$ needs to act trivially in $V$ for $\mathrm{Hom}(W, W \otimes V)$ to ever be non-trivial. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)

I don't know if this is also sufficient. This boils down to looking at groups with trivial center.

In any case, you originally asked about finite subgroups of SU(2) and its fundamental representation. These are all classified, and none of them have trivial center, so all satisfy this property. Proof of the last bit: any subgroup $G$ of SU(2) gives a subgroup $\overline{G}$ of SO(3) by projection. If $\overline{G}$ has even order, it has an element of order 2, which necessarily lifts to an element of order 4 whose square is $-1$, so $-1 \in G$. The only subgroups of SO(3) of odd order are the odd cyclic groups, which lift to Abelian groups. Some references are is these notes by Dolgachev or an earlier MO question.

One necessary condition (that suffices in your SU(2) case, and all others I could think of) is that the center of $G$ needs to act trivially in $V$. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)

I don't know if this is also sufficient. This boils down to looking at groups with trivial center.

One necessary condition (that suffices in your SU(2) case, and all others I could think of) is that the center of $G$ needs to act trivially in $V$ for $\mathrm{Hom}(W, W \otimes V)$ to ever be non-trivial. The character of the center justs multiplies in a tensor product, and so we can't have a map from $W$ to $V \otimes W$ if $V$ has a non-trivial central character. (This also follows from Ben's observation above.)

I don't know if this is also sufficient. This boils down to looking at groups with trivial center.