Does conjugation imply equivariantly conjugation? Let $G$ be a finite group acting on a finite dimensional vector space $V$. Let $A$, $B\in End_G(V)$ be two equivariant endomorphisms, i.e.
$$
gAg^{-1}=A, ~gBg^{-1}=B, \forall g \in G.
$$
Now suppose that $A$ and $B$ are conjugate, i.e. there exists a $J\in Gl(V)$ such that
$$
JAJ^{-1}=B.
$$
Then is it true that $A$ and $B$ are $G$-equivariantly conjugate? More precisely, can we find a $\tilde{J}\in Gl(V)$ such that $\tilde{J}$ is also $G$-equivariant and
$$
\tilde{J}A\tilde{J}^{-1}=B?
$$
A natural candidate of  $\tilde{J}$ is $\frac{\sum_{g\in G}gJg^{-1}}{|G|}$. But I cannot prove it is invertible.
If the statement is not true, is there any counter-example?
 A: 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.
