The following example shows that your bound is close to being optimal. Let $X$ be the product $$ \mathbb{P}^1 \times \mathbb{P}^1 \times \ldots \times \mathbb{P}^1,$$ of dimension $n$, and let $f$ permute the factors by a cyclic permutation. This extends to a automorphism of the ambient projective spaces for the embedding assosiated to $\mathcal{O}(1) \otimes \ldots \otimes \mathcal{O}(1)$.

Consider an $f$-equivariant embedding of $Y$. The embedding is given by some $\mathcal{O}(a_1) \otimes \ldots \otimes \mathcal{O}(a_k)$ acting by $f$ on the image and pulling back shows that $a_1=a_{2} = \ldots = a_{k}$. The dimension of the space of sections of $\mathcal{O}(1) \otimes \ldots \otimes \mathcal{O}(1)$ is $2^n$.

This was example was motivated by Remark 3.4.6 of "Linearization of algebraic group actions" by Brion.

The following example shows that there is a difference between equivariant finite maps and equivariant embeddings, there is no bound in terms of $n$ which garuantees an equivariant embedding. The idea of the example gives the following byproduct: If $f$ has finite order and there are $m$ subgroups appearing as the isotropy subgroup of some point(see below) then, the dimension of the projective space has to be at least $log_{2}(m) -1$.

For a space with a $\mathbb{Z}_k$-action, to every point there is a subgroup of $\mathbb{Z}_k$ we can assosiate, consisting of the elements fixing the point, called the isotropy subgroup. The argument uses a certain fact: Suppose that $\mathbb{Z}_k$ acts linearly on $\mathbb{CP}^n$, then the possiblities for the isotropy subgroup is at most $2^{n+1}$. To prove this note it is possible to lift to a linear transformation of $\mathbb{C}^{n+1}$ of order $k$, thus to diagonalize the lift. Then the isotropy subgroup of an element $p = [z_{0}: \ldots : z_{n}] \in \mathbb{CP}^n$ depends only on which of the $z_{i}$ are zero, and which are not zero. Therefore there are only $2^{n+1}$ possibilities. To make an analogy with toric geometry, the non-trivial possibilities for the strata are in bijecton with subsimplices of the $(n+1)$-simplex.

Consider the $\mathbb{C}^{*}$-action $\mathbb{CP}^{2}$ coming from the linear action on $\mathbb{C}^3$ with weights $(0,1,2)$. To any smooth invariant curve there is an isotropy subgroup defined similarly to the above, i.e. the isotropy subgroup of any point in the curve that is not fixed. After intersecting with $S^1$, we get a subgroup of the form $\mathbb{Z}_{n}$ (by using continuity of the action in the complex topology). In slight abuse of notation I will always intersect with $S^1$ when referring to the isotropy subgroup. 

The example is produced by doing a certain sequence of equivariant blow-ups in fixed points. i.e. equivariant blow-up of a fixed point, and then of a fixed point in the exeptional divisor, and so on. With this process it is possible to construct an action with isotropy subgroups $\mathbb{Z}_k$ for all $k=2,\ldots,k_0$, for any positive integer $k_{0}$. It follows that the restricted action of $\mathbb{Z}_{k_{0}} \subset \mathbb{C}^*$ can have arbitrarily large number of subgroups appearing as isotropy subgroups. These actions are linearizable, because the $\mathbb{C}^*$ action is. Since $H^{1}(S,\mathcal{O}_S)=0$, two line bundles with the same $c_1$ are isomorphic. Pick an ample line bundle, because the torus is connected, $g^*L \cong L$ for any $g \in \mathbb{C}^*$. It follows that the action is linearizable.

So, considering $f$ generating such an action on the surface $S$. Suppose we have an equivariant embedding $i: S \rightarrow \mathbb{CP}^{N}$. Consider the minimal projective subspace containing $i(S)$, it is invariant by a contradiction argument (you could move it by $f$ and intersect again). Furthermore, $f$ generates a faithful, linear $\mathbb{Z}_{k_0}$-action on this minimal subspace $\mathbb{CP}^n$, with $n < N$, by a similar argument. By the fact mentioned in the beginning, the more blow-ups we do to get $S$, the higher $n$ needs to be.