Let $ \overline{\mathbb{Q}}_{p} $ be an algebraic closure of $ p $-adic numbers $ \mathbb{Q}_{p} $. A closed subgroup $ H $ of a general linear group $ {\rm GL}_{n}(\overline{\mathbb{Q}}_{p}) $ is called completely reducible if the inclusion map $ i:H\to {\rm GL}_{n}(\overline{\mathbb{Q}}_{p}) $ is a completely reducible representation of $ H $.

Now let $ G $ be a topologically finitely generated closed subgroup of $ {\rm GL}_{n}(\overline{\mathbb{Q}}_{p}) $ and assume that $ G $ is completely reducible. Suppose that $ H $ is a closed subgroup of $ G $ and the union of the conjugacy classes of $ H $ in $ G $ is closed. My question is the following:

Is $ H $ also completely reducible?

For example, if $ H $ is a normal subgroup of $ G $, then it's true. Moreover, if we replace the above $ p $-adic topology with Zariski topology, then the question has a affirmative answer, cf. Theorem 4.1 in ALGEBRAIC GROUPS AND G-COMPLETE REDUCIBILITY: A GEOMETRIC APPROACH by BENJAMIN MARTIN. Thanks in advance.

$\DeclareMathOperator\GL{GL}$Let $ \overline{\mathbb{Q}}_{p} $ be an algebraic closure of $ p $-adic numbers $ \mathbb{Q}_{p} $. A closed subgroup $ H $ of a general linear group $ \GL_{n}(\overline{\mathbb{Q}}_{p}) $ is called completely reducible if the inclusion map $ i:H\to \GL_{n}(\overline{\mathbb{Q}}_{p}) $ is a completely reducible representation of $ H $.

Now let $ G $ be a topologically finitely generated closed subgroup of $ \GL_{n}(\overline{\mathbb{Q}}_{p}) $ and assume that $ G $ is completely reducible. Suppose that $ H $ is a closed subgroup of $ G $ and the union of the conjugacy classes of $ H $ in $ G $ is closed. My question is the following:

Is $ H $ also completely reducible?

For example, if $ H $ is a normal subgroup of $ G $, then it's true. Moreover, if we replace the above $ p $-adic topology with Zariski topology, then the question has a affirmative answer, cf. Theorem 4.1 in Algebraic groups and $G$-complete reducibility: A geometric approach by Benjamin Martin.