Trivial representation in Clebsch-Gordan decomposition

My professor defined the Clebsch-Gordan series as the direct sum decomposition of the tensor product of two representations of the Lie group SU(2):

$$D_{j_1} \otimes D_{j_2} = D_{j_1+j_2} \oplus D_{j_1+j_2-1} \oplus ... \oplus D_{|j_1-j_2|}$$

I think it's an interesting fact that the trivial representation $D_0$ will be in the tensor product if and only if $j_1=j_2$.

Is there an intuitive explanation for that? What common property of all systems with $j_1=j_2$ leads to the trivial representation being part of the tensor product?

This question could be generalized to a similar fact about $D_{1/2}$ and $D_1$.

Generically, given a representation $V$ of a group, the tensor representation $V\otimes V$ will decompose into the symmetric and antisymmetric parts $$V\otimes V = \Lambda^2 V \oplus S^2 V$$ and in the case of the rotation group (or its universal cover), the symmetric 2-tensors have a certain invariant under rotations - their trace! So when $j_1 = j_2$, the tensor representation is really the 2-tensors (think: matrices) on the $D_{j_1}$ space, and the trace part of the matrices forms the trivial subrepresentation.
If $j_1 \neq j_2$, the tensor representation is not simply such sqaure matrices on the original space, so we can't speak of a trace, and we don't have a trivial subrepresentation.
Yes. To obtain a scalar (or the trivial representation) one must multiply a representation $$D^{\lambda}$$ by its conjugate $$D^{\lambda^*}$$: the weights in the conjugate representation $$\lambda^*$$ are the negative of those in $$\lambda$$, and the scalar occurs when combining conjugate functions, so,as to produce a final function with weight $$0$$.
In $$SU(2)$$ all representations are self conjugate, so $$\lambda^*$$ is equivalent to $$\lambda$$, and indeed the weights in $$\lambda$$ runs from $$-\lambda$$ to $$\lambda$$, so that any weight and its negative are in the representation. This means that, to get a scalar, you need to have $$\lambda_1=\lambda_2$$.
In other groups, v.g. $$SU(3)$$, not all irreps are self-conjugate. To get a scalar one must then combine two properly conjugate irreps, such as $$(1,0)$$ and $$(0,1)$$. Indeed $$(1,0)\otimes (0,1)=(0,0)\oplus (1,1)$$ etc.