Yes, there is always such an $M$.

To see this, first note that saying two representations of $G$ on a vector space $V$ are equivalent is the same as saying the two images of $G$ in $Gl(V)$ are conjugate by an element of $Gl(V)$. A theorem due to Mal'cev which can be found in

Mal'cev. On semisimple subgroups of lie groups. Amer. Math. Soc. Trans., 1:172-273, 1950.

proves that if $V$ is given a Hermitian inner product and both images of $G$ lie in $U(V)\subseteq Gl(V)$, then they are equivalent representations iff the images are conjugate by an element of $U(V)$. (This is also true replacing $U(n)$ by any odd dimensional orthogonal group, $SU(n)$, or $Sp(n)$, but needs to be modified slightly in the case of even dimensional orthogonal groups.)

So, if each of the groups in your list has a unique nontrivial $2n$-dimensional representation, we'll be done. Unfortunately, this is true of $Sp(n)$ only. The group $Sp(n)\times S^1$ has infinitely many (given as the tensor product of the standard representation with any nontrival 1-d rep of $S^1$). On the other hand, $SU(2n)$ has precisely two given by the standard representation or the complex conjugate of the standard representation. Finally, $U(2n)$ has two infinite familes given on the cover $SU(2n)\times S^1$ as a tensor product of either of $SU(2n)$s reps with any nontrivial $1$-dim rep of $S^1$.

Let me just focus on the case of $Sp(n)\times S^1$, the other cases being similar. Consider the tensor product of the standard $2n$-dimensional rep of $Sp(n)$ together with any nontrivial $1$-dimensional rep of $S^1$, say, of weight $k$. By precomposing with the cover $\pi:Sp(n)\times S^1\rightarrow Sp(n)\times S^1$ where $\pi(A,z) = (A,z^k)$, we see that the image of $Sp(n)\times S^1$ in $U(2n)$ under this representation is the same as if we use the weight $1$ (standard) rep of $S^1$. Now, Mal'cev's theorem implies that the image of $Sp(n)\times S^1$ is conjugate to the standard $Sp(n)\times S^1$.