= Solution
Put $w_j=|C_j|/|G|$. The <character orthogonality> for the five supplied rows say $\sum_jw_j\chi_r(C_j)\chi_s(C_j)=\delta_{rs}$. Solving these linear equations gives
$$
(w_1,\ldots,w_7)=\frac1{120}(1,15,20,24,10,20,30).
$$
For example orthogonality of the two degree-four rows gives $16w_1+w_3+w_4-4w_5-w_6=0$, and their norms give $16w_1+w_3+w_4+4w_5+w_6=1$; the remaining row products determine the other weights. Since $C_1=\{e\}$, $w_1=1/|G|$, so \b[$|G|=120$] and the class sizes are the displayed numerators.
Tensoring the fifth <irreducible representation> with the second, one-dimensional one gives another <irreducible character>, with row $(5,1,-1,0,-1,-1,1)$. It differs from all five known rows. There are seven <conjugacy classes>, hence seven <irreducible characters>. The sum of squared degrees is $|G|$, so the final degree is $\sqrt{120-(1+1+16+16+25+25)}=6$.
The <regular representation> has character zero off the identity and is the sum of irreducible characters weighted by their degrees. Applying this column by column determines the last row. The completed <character table>, in the supplied class order, is
$$
\boxed{\begin{array}{c|rrrrrrr}
|C_j|&1&15&20&24&10&20&30\\\hline
\chi_1&1&1&1&1&1&1&1\\
\chi_2&1&1&1&1&-1&-1&-1\\
\chi_3&4&0&1&-1&2&-1&0\\
\chi_4&4&0&1&-1&-2&1&0\\
\chi_5&5&1&-1&0&1&1&-1\\
\chi_6&5&1&-1&0&-1&-1&1\\
\chi_7&6&-2&0&1&0&0&0
\end{array}.}
$$
No identification of the group with a familiar group is needed.
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