Is the dual representation of an irreducible representation always irreducible?

Let $G$ be a group and let $V$ be a complex vector space which is a representation of $G$. Let's write the (left) action of $g\in G$ on $v\in V$ as $gv$.

The dual vector space of $V$ is the set of linear maps from $V$ to $\mathbb{C}$, and is written as $V^*$. I'll use the notation $(\phi,v)$ to mean $\phi(v)$ if $\phi\in V^*$ and $v\in V$.

The dual representation of $V$ is defined to be the vector space $V^*$ with the (right) $G$-action $(\phi g,v):=(\phi,gv)$ for $\phi\in V^*$, $v\in V$ and $g\in G$.

A representation of $G$ is irreducible if the only $G$-invariant subspaces are the zero subspace or the whole vector space.

I can show that if $V^*$ is an irreducible representation, then so is $V$. Indeed, if $W$ is a $G$-invariant subspace of $V$ then its annihilator $W^\perp=\{\phi\in V^*\colon w\in W\implies (\phi,w)=0\}$ is a $G$-invariant subspace of $V^*$, so either $W^\perp=V^*$, which clearly implies $W=0$, or $W^\perp=0$, which (by fiddling around with a Hamel basis) yields $W=V$.

Does the converse hold? That is, if $V$ is irreducible, must $V^*$ also be irreducible?

If $\dim V<\infty$ then $V^{**}$ is equivalent to $V$, so the answer is yes in this case. But I don't know what happens if $\dim V=\infty$.

No. Let $G = S_{\infty}$ denote the group of permutations of $\mathbb{N}$ which fix all but finitely many elements. $G$ is countable, so any irreducible representation has at most countable dimension. $G$ has a countable-dimensional irreducible representation $V$ given by the subspace of $\bigoplus_{i=1}^{\infty} \mathbb{C}$ of sequences adding to $0$, and $V^{\ast}$ is of uncountable dimension, so cannot be irreducible. (Explicitly, there is an obvious bilinear pairing on $V$ giving $V^{\ast}$ a proper invariant subspace isomorphic to $V$.)

Edit: Here's a cute non-constructive argument. Suppose that $G$ has an infinite-dimensional irreducible representation $V$. If $V^{\ast}$ is reducible, we're done. Otherwise, $V$ is a proper invariant subspace of $V^{\ast \ast}$ (at least given the axiom of choice), so $V^{\ast}$ is an irreducible representation whose dual is not irreducible.