Yes, it is just $\operatorname{Sym}^k(V)$ itself.

Specifically: Let $V = \mathbb{C}^{2n}$ be the natural module for $Sp(2n, \mathbb{C})$. Then for all $1 \leq k \leq n$, the symmetric power $\operatorname{Sym}^k(V)$ is an irreducible $Sp(2n,\mathbb{C})$-module of highest weight $(k,0,\ldots,0) = k \varpi_1$.

See for example §24.2, p. 406 in Fulton-Harris.

Yes, it is just $\operatorname{Sym}^k(V)$ itself.

Specifically: Let $V = \mathbb{C}^{2n}$ be the natural module for $Sp(2n, \mathbb{C})$. Then for all $k \geq 1$, the symmetric power $\operatorname{Sym}^k(V)$ is an irreducible $Sp(2n,\mathbb{C})$-module of highest weight $(k,0,\ldots,0) = k \varpi_1$.

See for example §24.2, p. 406 in Fulton-Harris.

Some alternative references, which are also relevant when you look at the same question for $Sp(2n,K)$ with $K$ an algebraically closed field of positive characteristic.

• II.2.17 in: Jantzen, Jens Carsten Representations of algebraic groups. Second edition. Mathematical Surveys and Monographs, 107. American Mathematical Society, Providence, RI, 2003. xiv+576 pp. ISBN: 0-8218-3527-0.
• 8.1(c) in: Seitz, Gary M. The maximal subgroups of classical algebraic groups. Mem. Amer. Math. Soc. 67 (1987), no. 365, iv+286 pp.
• Suprunenko, I. D. Conditions for the irreducibility of the restrictions of irreducible representations of the group SL(n,K) to connected algebraic subgroups. Dokl. Akad. Nauk BSSR 30 (1986), no. 3, 204-207, 284.