Take a compact homogeneous space $G/K$, and a left $G$invariant differential $k$form $\omega \in \Omega^k(G/K)$. Will $\omega$ necessarily be closed? Might it even be harmonic when $G/K$ is endowed with a Riemannian metric?

6$\begingroup$ No. See Bredon's book "geometry and topology" for a good discussion of the complex of invariant differential forms. For $K = 1$ you recover the ChevalleyEilenberg complex of $\mathfrak g$. $\endgroup$ – Mike Miller Aug 13 '20 at 16:11

5$\begingroup$ They don't have to be closed, but each closed form differs from a closed invariant form by an exact (not necessarily invariant) form, so the de Rham cohomology of forms is the same as that of invariant forms, assuming $G$ is compact. $\endgroup$ – Ben McKay Aug 13 '20 at 16:59
Note that the answer depends on the pair $(G,K)$.
For example, if $K=\{e\}$, then one is asking whether the ring of leftinvariant forms on $G$ consists only of closed forms. This only happens when $G$ is abelian.
On the other hand, if $M=G/K$ is a compact Riemannian symmetric space and $G$ is the identity component of the isometry group of $M$, then, indeed, every $G$invariant form is closed and, in fact, the ring of $G$invariant forms on $M$ is equal to the space of harmonic forms on $M$. This is a wellknown result, but for a short proof, one can consult this note by Michael E. Taylor.
For example, when $M=\mathbb{CP}^n$ endowed with its FubiniStudy metric, one has $G = \mathrm{SU}(n{+}1)/\mathbb{Z}_{n+1}$, and the only $G$invariant forms are (linear combinations of) powers of the Kähler form $\omega$.
As another example, if $K$ is compact and $M = (K\times K)/\Delta$, where $\Delta = \{ (k,k)\ \ k\in K \}$, then the $(K\times K)$invariant forms on $M$ are simply the biinvariant forms on $K$, which are all closed.