Let $q=p^f$ be a prime power such that $q \equiv 1 \pmod 5$. According to the list of irreducible (complex) character degrees of $SL(5, q)$ in Frank Luebeck's homepage, $SL(5, q)$ has 20 irreducible characters of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. Using Lusztig's parametrization of irredicible complex characters of groups of Lie type, I want to prove the existence of such an irreducible character of $SL(5, q)$:

Let $\chi \in \mathrm{Irr}(SL(5, q))$ be of degree $1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$. It is known that if $p$ is sufficiently large (I think $p>5$ would be sufficient), then all the $p'$-degree irreducible characters of $SL(5, q)$ are preciesly the so-called semisimple characters which are corresponding to the semisimple conjugacy classes of $PGL(5, q)$. Therefore, $\chi=\chi_s$ where $s\in PGL(5, q)$ is a semisimple element and $\chi(1)=\displaystyle\frac{|SL(5, q)|_{p'}}{|C_{PGL(5, q)}(s)|_{p'}}$.

Question: I wonder $\chi$ corresponds to which semisimple element $s$ of $PGL(5, q)$?

Some Thoughts: We deduce from above argument that we must have $|C_{PGL(5, q)}(s)|_{p'}=5(1+q+q^2+q^3+q^4)$. So if $s_1$ is a preimage of $s$ in $GL(5, q)$, the characteristic polynomial of $s_1$ is an irreducible polynomial of degree $5$ in $\mathbb{F}_q[x]$. Hence we have $C_{GL(5, q)}(s_1)\cong GL(1, q^5)$ and $C_{PGL(5, q)}(s)\cong C_{GL(5, q)}(s_1)/Z(GL(5, q))$. But this implies that $|C_{PGL(5, q)}(s)|_{p'}=1+q+q^2+q^3+q^4$ and so $SL(5, q)$ could not contain an irreducible character of degree$1/5(q^2+1)(q^2+q+1)(q+1)^2(q-1)^4$! How can I explain this?
I would be grateful if you could hint me how to find a proper semisimple element corresponds to $\chi$.