Is an equivariant projective morphism equivariantly-projective? Let everything be over $\mathbb{C}$. Consider two varieties $X,$ $Y,$ where $X$ is normal and $Y$ is affine,
having regular $\mathbb{C}^*$-actions and
a $\mathbb{C}^*$-equivariant projective morphism
$\pi: X \rightarrow Y.$ The "projective" means that it factors trough an closed immersion and projection,
$$\pi=\pi_Y \circ \iota : X \hookrightarrow \mathbb{P}^n \times Y \rightarrow  Y$$
for some $n.$
Now, the question: Can this immersion be made $\mathbb{C}^*$-equivariant? That is, having a linear $\mathbb{C}^*$-action
on $\mathbb{P} ^ M$ for some $M$ such that $\iota$ is $\mathbb{C}^*$-equivariant, where on $\mathbb{P} ^M \times Y$ we consider the diagonal action.
 A: No. Let $Y$ be a point and $X$ be $\mathbb P^1$ with two points glued together. Let $G=\mathbb G_m$ act on $X$ by its usual action on $\mathbb P^1$ fixing those two points, and acting trivially (of course) on $Y$.
Then this morphism is not equivariantly projective because, for any $\mathbb G_m$ action on $\mathbb P^M$, and any $x \in \mathbb P^M$ not fixed, the two limits of the $\mathbb G_m$-translates of $x$ as the parameter in $\mathbb G_m$ goes to $0$ and $\infty$ must be distinct.
One can check this by working in coordinates which are eigenvectors for the $\mathbb G_m$ action - one limit involves taking the nonvanishing coordinates of $x$ where $\mathbb G_m$ acts by the greatest power and one by taking the nonvanishing coordinates of $x$ where $\mathbb G_m$ acts by the least power, and these are only equal if $\mathbb G_m$ acts by only a single power on nonvanishing coordinates of $x$, in which case $x$ is fixed by $\mathbb G_m$.
Conversely, for any smooth point on $x$, the two limits of the $\mathbb G_m$-translates are the same, nodal, point.
So you will at least need to assume normalcy or something on $X$ and not just $Y$.
