This seems to be completely answered in the topological category by Thm 1.4 of http://arxiv.org/pdf/1202.3368v3.pdf.

Edit

If X is a topological space which is not Dieudonne complete (meaning its topology cannot be given by a complete uniformity), then it seems theorem 7.24 of http://books.google.com/books?id=v3_PVdvJek4C&pg=PA35&lpg=PA35&dq=free+topological+group+dieudonne+complete&source=bl&ots=QIt0C2TCjN&sig=d4BJjO2h3zM4jDLoTzijrDVYt3w&hl=en&sa=X&ei=VIw8UeuBJpO30QHek4C4AQ&ved=0CC4Q6AEwAA and Thm 1.4 of the paper above shows that the free topological group on X is not the isometry group of a metric space. Googling shows completely regular spaces exist which are not Dieudonne.

I believe that any polish group or locally compact group is an isometry group of a metric space by the paper I linked.

Also the author of the first paper has shown that every Lie group is the isometry group of another Lie group with respect to some proper metric.

This seems to be completely answered in the topological category by Thm 1.4 of http://arxiv.org/pdf/1202.3368v3.pdf.

Edit

If X is a topological space which is not Dieudonne complete (meaning its topology cannot be given by a complete uniformity), then it seems theorem 7.24 of http://books.google.com/books?id=v3_PVdvJek4C&pg=PA35&lpg=PA35&dq=free+topological+group+dieudonne+complete&source=bl&ots=QIt0C2TCjN&sig=d4BJjO2h3zM4jDLoTzijrDVYt3w&hl=en&sa=X&ei=VIw8UeuBJpO30QHek4C4AQ&ved=0CC4Q6AEwAA and Thm 1.4 of the paper above shows that the free topological group on X is not the isometry group of a metric space. Googling shows completely regular spaces exist which are not Dieudonne.

I believe that any polish group or locally compact group is an isometry group of a metric space by the paper I linked.

Also the author of the first paper has shown that every Lie group is the isometry group of another Lie group with respect to some proper metric. http://arxiv.org/pdf/1201.5675v2.pdf

This seems to be completely answered in the topological category by Thm 1.4 of http://arxiv.org/pdf/1202.3368v3.pdf.

This seems to be completely answered in the topological category by Thm 1.4 of http://arxiv.org/pdf/1202.3368v3.pdf.

Edit

If X is a topological space which is not Dieudonne complete (meaning its topology cannot be given by a complete uniformity), then it seems theorem 7.24 of http://books.google.com/books?id=v3_PVdvJek4C&pg=PA35&lpg=PA35&dq=free+topological+group+dieudonne+complete&source=bl&ots=QIt0C2TCjN&sig=d4BJjO2h3zM4jDLoTzijrDVYt3w&hl=en&sa=X&ei=VIw8UeuBJpO30QHek4C4AQ&ved=0CC4Q6AEwAA and Thm 1.4 of the paper above shows that the free topological group on X is not the isometry group of a metric space. Googling shows completely regular spaces exist which are not Dieudonne.

I believe that any polish group or locally compact group is an isometry group of a metric space by the paper I linked.

Also the author of the first paper has shown that every Lie group is the isometry group of another Lie group with respect to some proper metric.