This is just an expanded version of Igor's comment. Indeed this is an open problem. The common opinion (I believe) is that such groups do exist, but the best result in this direction so far is the Olshanskii-Sapir group, which is finitely presented and (infinite torsion)-by-cyclic.
There is a general idea, commonly attributed to Rips, which shows that such groups should exist. The idea is the following. Take the free Burnside group $B(m,n)$ on $m\ge 2$ generators and of exponent $n>>1$, so that $B(m,n)$ is infinite. Embed it into a finitely presented group $G$ by using a version of the Higman embedding theorem. Let $G=\langle x_1, \ldots , x_n\rangle$. Then take the quotient group $Q$ of $G$ by the relations
(*) $x_1=b_1, \ldots , x_n=b_n$, where $b_1, \ldots , b_n\in B(m,n)$.
Clearly $Q$ is torsion being a quotient of $B(m,n)$ and is finitely presented. And it is believable that if the embedding of $B(m,n)$ in $G$ is "reasonably hyperbolic" and elements $b_1, \ldots , b_n$ are chosen randomly, then $Q$ is infinite with probability close to $1$. This idea was essentially implemented by Olshanskii and Sapir, but they managed to impose all relations of type (*) but one. (This is in fact a very rough interpretation of what they did, so Mark may correct me here.) There are even very particular choices of $b_1, \ldots , b_n$ (e.g., long aperiodic words with small cancellation), for which the group $Q$ should be infinite. So there are even particular finite presentations which should represent infinite torsion groups, but nobody knows how to prove that.
Theorem B of this paper implies that we can take any two nontrivial involution-free groups $A$ and $B$ and construct a
complete simple groups $D$ with a (diagrammatically) aspherical presentation $D=A*B/\langle\!\langle w_1, w_2,\dots\rangle\!\rangle$ (though such use of this theorem is killing a mosquito with a cannon).
Here, complete means naturally isomorphic to the automorphism group (i.e. centreless and without outer automorphisms). Now, suppose that $A$ and $B$ coincides with their commutator subgroups.
Acphericity implies that the centre of the free central extension $\widetilde D=A*B/[\langle\!\langle w_1, w_2,\dots\rangle\!\rangle,A*B]$ of $D$ is the free abelian group with the basis $\widetilde{w_1},\widetilde{w_2},\dots$ (see, e.g., Olshanskii's book, Section 34.4). Now, we can do whatever we want. For instance, we can take the quotient $G=\widetilde D/\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ and obtain a desired group $G$.
The natural map $\mbox{Aut}\,\widetilde D\to\mbox{Aut}\,G$ exists because the subgroup $\langle\widetilde {w_1}^p,\widetilde{w_2}^p,\dots\rangle$ is characteristic in $\widetilde D$ (as this subgroup consists of $p$th powers of all central elements), and this map is injective because
$\widetilde D$ coincides with its commutator subgroup. So, there are no outer automorphisms of $G$.
Jul 6, 2015. More details added, as suggested by Mamuka.
(1)
$G$ (as well as $D$ and $\widetilde D$) is finitely generated if $A$ and $B$ are finitely generated.
(2)
Suppose that a group is a quotient of another group: $L=M/N$. Then
there is a natural map $f\colon \mbox{Aut}\,M\to \mbox{Aut}\,L$ if $N$ is characteristic;
$f(\mbox{Inn}\,M)=\mbox{Inn}\,L$ (i.e. $f$ sends inner automorphisms to inner automorphisms, and each inner automorphism of $L$ has at least one inner preimage);
$f$ is injective if $N$ is central and $M$ coincides with its commutator subgroup.
Now, take $M=G$ and $L=D$ (and $N=\langle\widetilde {w_1},\widetilde{w_2},\dots\rangle$). Then $\mbox{Aut}\,L=\mbox{Inn}\,L$ and, hence,
$\mbox{Aut}\,M=\mbox{Inn}\,M$ by (2), i.e.
all automorphisms of $M=G$ are inner.
Best Answer
I am pretty sure this example can be constructed. It might be not easy, though, since you need to have finite order automorphisms of the Tarski monsters without non-trivial fixed points. I do not remember if that has been done for Tarski monsters. For other monsters it was done, I think, by Ashot Minasyan in Groups with finitely many conjugacy classes and their automorphisms. Ashot is sometimes on MO, perhaps he can give more details. If not, you can ask him directly.
I talked to Olshanskii and Osin today about this problem. They confirmed the answer. Moreover, they reminded me that there exists a paper Obraztsov, Viatcheslav N. Embedding into groups with well-described lattices of subgroups. (English summary) Bull. Austral. Math. Soc. 54 (1996), no. 2, 221–240 where he constructs Tarski monsters with arbitrary prescribed outer automorphism group. It is not difficult to deduce from the construction that if the prescribed outer automorphism group is, say, of order 3, then it does not have non-trivial fixed points which gives the property from the question.
Another, more straightforward, way to construct an example is the following. Fix the free group $F_2$ and its obvious automorphism $\alpha$ of order 2 that switches the generators. Now consider the procedure of creating the Tarski monster quotient of $F$ described in the book by Olshanskii "Geometry of defining relations...". At each step together with a new relation $r_i=1$, impose the relation $\alpha(r_i)=1$. It is not difficult to see that $\alpha$ will be an outer automorphism of the resulting Tarski monster without non-trivial fixed points.