Does conjugation imply equivariantly conjugation?

Consider $G = S_3$ and let $V = 1 \oplus \text{sgn}$ be the direct sum of the trivial and sign representations.

Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$, then these are both $G$-equivariant endomorphisms of $V$ conjugate by any $J = \begin{pmatrix} 0 & a \\ b & 0 \end{pmatrix}$ where $a,b \in \mathbb{C}^*$. These are the only possible change of basis matrices.

Now, for any $\sigma \in S_3$ such that sgn($\sigma) = -1$ we have that $\sigma$ acts as $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ on $V$, by abuse of notation label this matrix $\sigma$, then

$$\sigma J \sigma^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 0 & a \\ b & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & -a \\ -b & 0 \end{pmatrix} = -J.$$ So $J$ is not $S_3$-equivariant for any $a, b$.

In fact, by Schur's Lemma, $\text{dim}_G(V,V) = \text{dim}_G(1 \oplus \text{sgn},1 \oplus \text{sgn}) = 2$ so any $G$-equivariants are linear combinations of $A$ and $B$ above, so $P$ never stood a chance.