Centralizers of semisimple subgroups $\DeclareMathOperator\GL{GL}$If $G$ is a simple Lie group, and $\rho: G \to \GL(V)$ is a representation, then by Schur's lemma, the group of automorphisms of $\rho$ is a reductive subgroup of $\GL(V)$. I'm wondering whether this generalizes to the case where $\GL(V)$ is replaced by an arbitrary reductive group?
More generally: if $G$ is a semisimple algebraic group (or even reductive), $H$ is a reductive algebraic group, and $\rho : G \to H$ is a homomorphism, is the centralizer of the image of $\rho$ in $H$, $C_{H}(\rho(G))$, a reductive subgroup of $H$?
Edits:

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*The case I'm most interested is when the field is $\mathbb{C}$, but I would be interested in hearing about other cases as well.


*As pointed out in the comments, in the case $G = H$, the centralizer is the centre, which is reductive but not necessarily connected. So the centralizer may not be connected in the general case as well.
 A: In characteristic zero, the connected closed subgroups of $G$ are in 1-1 correspondence with the Lie subalgebras of ${\rm Lie}(G)={\mathfrak g}$, and the Killing form a suitably chosen symmetric bilinear $G$-equivariant form on ${\mathfrak g}$ is non-degenerate. In fact we can choose this form to be the trace form for a rational representation for $G$. Since the Killing form this form is $G$-equivariant, it is $H$-equivariant, and irreducible summands for non-isomorphic $H$-submodules are orthogonal. Hence the Killing form on ${\mathfrak g}$ restricts to a non-degenerate form on ${\mathfrak g}^H={\rm Lie}(Z_G(H)^\circ)$.
I now claim that if $K$ is a connected algebraic group with a non-degenerate trace form $\kappa$ arising from a rational representation $\rho:K\rightarrow {\rm GL}(V)$, then $K$ is reductive. Indeed, if the unipotent radical $R_u(K)$ is non-trivial then $\rho(R_u(K))$ is a unipotent subgroup of ${\rm GL}(V)$, so after conjugation is contained in the subgroup of upper-triangular unipotent matrices, so the restriction of $\kappa$ to ${\mathfrak u}={\rm Lie}(R_u(K))$ is zero. Since ${\mathfrak u}$ is an ideal of ${\rm Lie}(K)$, this contradicts the non-degeneracy of the form. In particular, $Z_G(H)^\circ$ is reductive.
