Assuming $p:X\rightarrow Y$ a cover, and $p^{-1}(y) = n <\infty$. Is it possible for $\Delta$ to include more transformations than $|p^{-1}(y)|$ means $|\Delta| > |p^{-1}(y)|$ ?.
Thoughts concerning the question: At first I was trying to prove that $|\Delta| \le |p^{-1}(y)|$, I know that in case $X$ is simply connected, $\Delta \cong \pi_1(X)\cong p^{-1}(y)$, but in case $X$ is not simply connected I couldn't find a justification for a general $X$. What made me feel it might be wrong is the fact that each $D\in \Delta$ induces a permutation over $p^{-1}(y)$ which doesn't have fixed point (except for $D=id$). But this may result in more then $p^-{1}(y)$ permutations.
I found a solution:
Using the lifting property with $(X,x_0)\overset p \rightarrow (Y,y_0)$ and $(X,x_0) \overset p \rightarrow (Y,y_0)$, the uniqueness implies that the only lift $X\overset f\rightarrow X$ for $p$ ($p\circ f = p$) is id. Now lets us assume there are two deck transformations $D_1,D_2$ such that $D_1(x_0) = D_2(x_0) \equiv x_1$ (agree over a point of the fiber) then we may use the lifting property again with: $(X,x_1)\overset p \rightarrow (Y,y_0) , (X,x_1)\overset p \rightarrow (Y,y_0)$ and by the uniqueness we learn that $D_1=D_2$ because they both satisfies $p \circ D_i = p$.
So every pair $D_1,D_2\in D$ induces permutations which over $p^{-1}(y)$ which are "disjoint" to each other, means $D_1(x)\neq D_2(x)$ for every $x\in p^{-1}(y)$. Thus $|\Delta|\le|$ cyclic permutations over $p^{-1}(y)$ $| = |p^{-1}(y)|$.