Let $G$ be a group and let $\pi: G\rightarrow Aut_\mathbb{C}(V)$ be a finite dimensional irreducible representation of $G$.
I have two related questions
If I have two different hermitian inner products on $V$ with respect to which $\pi$ is unitary, does one have to be a scalar multiple of the other?
are two equivalent unitary finite dimensional irreducible representations of a group unitarily equivalent?
remark: (1. implies 2.)
Let $\left< \cdot, \cdot \right>_1, \left< \cdot, \cdot \right>_2$ be two Hermitian inner products on $V$ with respect to which $\pi$ is unitary and let $T \colon V \rightarrow V$ be the (unique) operator satisfying
$$ \left< Tv, w \right>_2 = \left< v, w \right>_1 $$
for all $v,w \in V$. Then
$$ \left< (\pi(g) \circ T)(v), v \right>_2 = \left< T(v), \pi(g)^{*}(v) \right>_2 = \left< T(v), \pi(g^{-1})v \right>_2 = \left< v, \pi(g^{-1})(v) \right>_1 = \left< v, \pi(g)^{*}(v) \right>_1 = \left< \pi(g)v, v \right>_1 = \left< (T \circ \pi(g))(v), v \right>_2 $$
for all $v \in V$ and $g \in G$ so $\pi(g) \circ T = T \circ \pi(g)$ for all $g \in G$ and $T$ is an intertwining operator. By Schur's lemma, $T = \lambda \cdot \operatorname{id}_V$ and since $T$ is $\left< \cdot, \cdot \right>_2$-self-adjoint (and positive) we have $\lambda \in \mathbb{R}$ and $\left< \cdot, \cdot \right>_2$ is a positive multiple of $\left< \cdot, \cdot \right>_1$.
The first one is surely false. Just take the trivial representation. Nevertheless, if $h_1,h_2$ are two invariant hermitian forms then $h_2(x,y)=h_1(Tx,y)$ for some invariant linear operator $T$, so by Schur's lemma they are proportional if your representation is irreducible.
Proof of the fact claimed in the proof: hermitian form $h$ is equivalent to an antilinear mapping $k$ from the vector space to its dual. You can take $T=k_2^{-1}k_1$.