Unitary matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are on the unit circle. Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are real. So unitary Hermitian matrices are precisely the matrices admitting a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $\pm 1$.
This is a very strong condition. As George Lowther says, any such matrix $M$ has the property that $P = \frac{M+1}{2}$ admits a complete set of orthonormal eigenvectors such that the corresponding eigenvalues are $0, 1$; thus $P$ is a Hermitian idempotent, or as George Lowther says an orthogonal projection. Of course such matrices are interesting and appear naturally in mathematics, but it seems to me that in general it's more natural to start from the idempotence condition.
I suppose one could say that Hermitian unitary matrices precisely describe unitary representations of the cyclic group $C_2$, but from this perspective the fact that such matrices happen to be Hermitian is an accident coming from the fact that $2$ is too small.
1) the set of diagonal matrices is path connected: if $A=\sum a_j E_{jj}$, $B=\sum b_j E_{jj}$ we take the map $t\mapsto \sum (ta_j+(1-t)b_j) E_{jj}$, $t\in[0,1]$.
2) The set of unitaries is path connected. If $U,V$ are two unitaries, we can always write them as $U=e^{iA}$, $V=e^{iB}$ with $A,B$ hermitian. Then we can consider the map $t\mapsto e^{itA}e^{i(1-t)B}$, $t\in[0,1]$ which gives a path from $U$ to $V$ within the unitary group.
3) The set of invertible hermitian matrices with positive eigenvalues is path connected. If $A,B$ are like that, then $A=UD_AU^*$, $B=VD_BV^*$. By parts 1) and 2), there exist continuous $f,g:[0,1]\to M_n(\mathbb{C})$ with $f(0)=D_A$, $f(1)=D_B$, $g(0)=U$, $g(1)=V$. Then $t\mapsto g(t)f(t)g(t)^*$ is continuous and takes $A$ to $B$. Note that the way that $f$ is defined guarantees that $f(t)$ will have positive eigenvalues for all $t\in[0,1]$.
4) GL$_n(\mathbb{C})$ is connected: Given $A,B$ invertible, we can write them as $A=RU$, $B=SV$ with $R,S$ hermitian and positive, and $U,V$ unitaries. By 3) and 2) we can find continuous functions $f,g:[0,1]\to M_n(\mathbb{C})$ with $f(0)=R$, $f(1)=S$, $g(0)=U$, $g(1)=V$. Then the map $t\mapsto f(t)g(t)$ is a continuous path from $A$ to $B$ (note that $f(t)$ and $g(t)$ are invertible for every $t\in[0,1]$ and then so is their product).
It only remains to justify the polar decomposition $A=RU$. An easy way to see this is by using the singular value decomposition. We write $A=WDV$, with $W,V$ unitaries and $D$ diagonal with non-negative entries (positive if $A$ is invertible). Then we can write
$$
A=(WDW^*)WV=RU,
$$
where $R=UDU^*$ is hermitian with positive eigenvalues (because $D$ is), and $U=WV$ is a unitary.
Best Answer
The eigenvalues of a Hermitian matrix are real (prove it). Thus $A+iE$ is invertible, because $-i$ is not an eigenvalue of $A$. But then $$ E-iA=-i(A+iE) $$ is invertible as well. Now compute \begin{align} \bigl((E+iA)(E-iA)^{-1}\bigr)\bigl((E+iA)(E-iA)^{-1}\bigr)^H &=(E+iA)(E-iA)^{-1}\bigl((E-iA)^{-1}\bigr)^H(E+iA)^H \\[6px] &=(E+iA)(E-iA)^{-1}(E+iA)^{-1}(E-iA) \\[4px] &=(E+iA)\bigl((E+iA)(E-iA)\bigr)^{-1}(E-iA) \\[4px] (\text{*}) &=(E+iA)\bigl((E-iA)(E+iA)\bigr)^{-1}(E-iA) \\[6px] &=(E+iA)(E+iA)^{-1}(E-iA)^{-1}(E-iA) \\[6px] &=E \end{align} The equality marked $(\text{*})$ is justified because $$ (E+iA)(E-iA)=E+A^2=(E-iA)(E+iA) $$
For the second case, you need to prove that $$ \bigl(-i(B-E)(B+E)^{-1}\bigr)^H=-i(B-E)(B+E)^{-1} $$ The left hand side can be rewritten as $$ i(B^H+E)^{-1}(B^H-E) $$ so you need to prove that $$ i(B^H+E)^{-1}(B^H-E)=-i(B-E)(B+E)^{-1} $$ which is equivalent to $$ (B^H-E)(B+E)=-(B^H+E)(B-E) $$ that only requires doing the operations on both sides. Justify each previous step.