Smooth map homotopic to Lie group homomorphism

If $G$ is a compact simply-connected simple Lie group, then any nontrivial homomorphism $G\to G$ is an automorphism (it is injective because $G$ is simple, and any immersion of closed connected manifolds of the same dimension is covering map), and in particular it has degree $\pm 1$. For example, if $f: S^3\to S^3$ is a map of degree $d$ with $|d|>1$, then $f$ is not homotopic to a homomorphism.

On the other hand, by obstruction theory any self-map of an $n$-torus is homotopic to a map induced by multiplication by an $n\times n$ matrix with integer entries, which is a homomorphism.

As Igor shows, every endomorphism of a simple Lie group $G$ has degree $\in\{0,\pm 1\}$.

On the other hand, every compact Lie group admits self maps of other degrees. Namely, the $k$-th power map $g\mapsto g^k$ has degree $k^r$, where $r$ is the rank of the group. So, each $k$ with $|k|\geq 2$ gives an example of a smooth map which is not homotopy equivalent to a homomorphism.

One way to compute the degree of the $k$-th power map is as follows. First, we can find an element $g\in G$ which lies in a unique maximal torus $T^r$ and which is also a regular value of the $k$-th power map. The uniqueness of the maximal torus implies that all $k$-th roots of $g$ lie in $T^r$, so this reduces the degree calculation to $T^r$, where it is obvious.

I’m not sure if this is what you’re looking for, but the map $\mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ given by sending $0\mapsto 1\,, 1\mapsto 0$ isn't homotopic to a homomorphism. More generally if the codomain is disconnected the answer to your question seems to be positive.