$\DeclareMathOperator\U{U}\DeclareMathOperator\SU{SU}$It's $\U(5) \times \U(1) \times \U(1)$.
We have a natural map from the normalizer of $G$ to the outer automorphism group of $G$. The outer automorphism group of $G$ is $\mathbb Z/2$, generated by the inverse transpose / complex conjugation.
This map is trivial, because if it were nontrivial, an element mapping to $1 \in \mathbb Z/2$ would give an isomorphism between this representation and the complex conjugate $\mathbf 5 \oplus \overline{\mathbf {10}} \oplus \mathbf 1$, which does not exist as these are not isomorphic.
The kernel of this natural map, for any $G \subseteq H$, is $G C_{H}(G)$, where $C_H(G)$ is the centralizer of $G$ in $H$.
Since every element of the kernel acts by conjugation as an inner automorphism, i.e. is an element of $G$ times something that acts trivially by conjugation, i.e. is an element of $G$ times something in its centralized, so the normalizer is $G C_H(G)$.
The centralizer is $\U(1) \times \U(1) \times \U(1)$ since there are three non-isomorphic irreducible representations of $G$, so the centralizer of $G$ consists of matrices acting by scalars on these three representations.
So the normalizer is the product, inside $\U(16)$, of this $\SU(5)$ and the $\U(1) \times \U(1) \times \U(1)$ scalars. We can check that this product is isomorphic to $\U(5) \times \U(1) \times \U(1)$ by letting $\U(5)$ act on $\overline{\mathbf 5}$ by the dual of the standard representation, $\mathbf{10}$ by the $\wedge^2$ of the standard representation, and $\mathbf 1$ trivially, and letting the two factors of $\U(1)$ act by multiplication on the last two representations.
